# Full text of "Elements of the Differential and Integral Calculus"

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About Google Book Search Google's mission is to organize the world's information and to make it universally accessible and useful. Google Book Search helps readers discover the world's books while helping authors and publishers reach new audiences. You can search through the full text of this book on the web at |http: //books .google .com/I k • MATHEMATICAL TEXTS Edited by Percey F. Smith, Ph.D. Professor of Mathematics in the Sheffield Scientific School of Yale University Elements of Differential and Integral Calculus By W. A. Granville, Ph.D. Elements of Analytic Geometry By P. F. Smith and A. S. Gale, Ph.D. Introduction to Analytic Geometry By P. F. Smith and A. S. Gale, Ph.D. Theoretical Meclianics In press By P. F. Smith and O. C. Lester, Ph.D. Advanced Algebra By H. £. Hawkes, Ph.D. Strength of Materials By S. E. Slocum, Ph.D., and E. L. Hancock, M.Sc. ELEMENTS OF THE DIFFERENTIAL AND INTEGRAL CALCULUS B7 WILLIAM ANTHONY GRANVILLE, Ph.D. nmTRUOTOiR or MATHBMATirS IK THE SHEFFIELD SCISNTIFIO SCHOOL Tale Univebsity With the Editorial Cooperation of PERCEY F. SMITH, Ph.D. pbofe880b of mathematics ik the sheffield soibktifio school Yale Ukivebsity f Dm ( A] |f_\" (if -^i, -^, -%>-. .t, ^^ ^^ .;j;^, ,.^^^ .. -, , ^ . , -.^ .J GINN & COMPANY BOSTON • NEW YORK • CHICAGO • LONDON \ . V I 623251 COPTRI^p, 1904, BY WnAAAM AlTTHONY QBAMVILLE AND PESCEY F. SMITB ALL RIGHTS BB8KBYXD 27.3 CINN & COMPANY • PRO- PRIETORS • BOSTON • U^Jk. PEEFACE The present volume is the result of an effort to write a modem text-book on the Calculus which shall be essentially a drill book. With this end in view, the pedagogic principle that each resiQt should be made intuitionally as well as analytically evident to the student has been kept constantly in mind. Indeed it has been thought best in some cases, as for example in Maxima and Minima and the Theorem of Mean Value, to discuss the question first from the intuitional side, in order that the significance of the new idea might be made plain in the ^ost direct manner. The object has not been to teach the student to rely upon his intuition^ but in some cases to use this facility in advance of the analytical investi- gation. The short chapter on Numbers is intended to give the student a chance to review his ideas of number. Limits and Continuity are treated at length, the latter mostly from a graphical standpoint, — the only method suited to a first course. In fact, graphical illustration has been di^wn upon to the fullest extent throughout the book. As special features, attention may be called to the effort to make perfectly clear the nature and extent of each new theorem, the large number of carefully graded exercises, and the summa- rizing into working rules of the methods of solving problems. In the Integral Calculus the notion of integration over a plane area has been much enlarged upon, and integration as the limit of a summation is constantly emphasized. The book contains more material than is necessary for the usual course of one hundred lessons given in our colleges and engineering schools; but this gives teachers an opportunity to choose such topics as best suit the needs of their classes. It is believed that the volume contains all subjects from which a selection naturally would be made in preparing students either for elementary work in applied sciencQ or for more advanced work in pure mathematiQs, iU iv PREFACE Certain proofs of considerable difficulty (as the existence of the number e) have been inserted with the belief that, while it is not always advisable to require beginners to learn them, a discussion of them with the class will render such investigations profitable and stimulating. With a few exceptions the author has found it impracticable to acknowledge his indebtedness to the large number of American, English, and continental writers whose books and articles have helped and inspired him in the work, the bulk of the matter having long been the common property of all mankind. While many of the exercises are new, a large number are standard and are to be found in many of the best treatises. The author's acknowledgments are due to Professor M. B. Porter of the University of Texas for critically examining the manuscript, to Professor James Pierpont of Yale University for many valuable suggestions, to my former colleagues. Professor E. R. Hedrick of the University of Missouri and Dr. C. N. Haskins, for their interest and assistance, to Dr. C. E. Stromquist of the University of Princeton for verifying the examples, and to my colleague, Mr. L. C. Weeks, for drawing the figures. The thanks of the author are also due to his former instructor in mathe- matics. Professor John E. Clark, now Professor Emeritus of Yale University, who first advised and encouraged him to undertake the task of writing this book. Sheffield Scientific School, Tale Univer8itt, July, 1904. CONTENTS DIFFERENTIAL CALCULUS Chapter I COLLECTION OF FORMULAS SEOnOK PAOX 1. Formulas for reference from Algebra, Trigonometry, and Analytic Geometry « 1 2. Greek alphabet 4 3. Natural values of trigonometric functions . * . . • 4 4. Rules for signs in the four quadrants 6 Chapter II NUMBERS 5. Rational numbers 7 6. Comparison of rational numbers with the points of a straight line . 7 7. Irrational numbers 8 8. Real numbers 8 9. Numerical or absolute value 10. Imaginary numbers 9 11. Complex numbers 9 12. Division by zero excluded 10 13. Only real numbers considered 10 Chapter III VARIABLES AND FUNCTIONS 14. Variables . 11 15. Constants 11 16. Interval of a variable 11 17. Continuous variation 11 18. Functions 12 19. Independent and dependent variables . . . . . .12 20. Notation of functions 13 21. Values of the independent variable for which a function is defined . 14 V vi CONTENTS SECTION PAGE 22. One-valued and many-yalued functions ...... 14 23. Explicit functions 15 24. Inyerse functions 15 25. Integral rational functions 16 26. Rational functions 16 27. Explicit algebraic functions 17 28. Transcendental functions 17 Chapter IV THEORY OF LIMITS 20. Limit of a variable 19 80. Infinitesimals 21 31. The concept of infinity (oo) 21 32. Limiting value of a function 22 33. Continuous and discontinuous functions 22 34. Continuity and discontinuity of functions illustrated by their graphs 23 35. Fundamental theorems on limits 26 36. Special limiting values .29 37. The number e .31 Chapter V DIFFERENTIATIOlf 38. Introduction 37 39. Increments 37 40. Comparison of increments 38 41. Derivative of a function of one variable 39 42. Symbols for derivatives 40 43. Differentiable functions 41 44. General rule for differentiation ....... 42 45. Applications of the derivative to Greometry 43 Chapter VI RULES FOR DIFFERENTIATING STANDARD ELEMENTARY FORMS 46. Importance of general rule 46 47. Differentiation of a constant 48 48. Differentiation of a variable Yiiih respect to itself .... 49 49. Differentiation of a sum 49 50. Differentiation of the product of a constant and a variable . . 50 CONTENTS vii BEOnOir PAGE 51. Differentiation of the product of two variables .... 50 52. Differentiation of the product of any finite number of variables . 50 53. Differentiation of a variable with a constant exponent ... 51 54. Differentiation of a quotient 52 55. Differentiation of a function of a function 57 56. Differentiation of inverse functions 58 57. Differentiation of a logarithm 58 58. Differentiation of the simple exponential function .... 60 59. Differentiation of the general exponential function ... 61 60. Logarithmic differentiation . . ' 62 61. Differentiation of sin v 67 62. Differentiation of coav 68 63. Differentiation of tan v 68 64. Differentiation of cot o 69 65. Differentiation of sec v 69 66. Differentiation of CSC . 70 67. Differentiation of vers v 70 68. Differentiation of arc sin v 74 69. Differentiation of arc cos <; 75 70. Differentiation of arc tan v 76 71. Differentiation of arc cot© 77 72. Differentiation of arc sec t; 77 73. Differentiation of arc esc t; 78 74. Differentiation of arc vers v 78 75. Implicit functions 83 76. Differentiation of implicit functions 84 Chapter VII SIMPLE APPLICATIONS OF THE DERIVATIVB 77. Direction of a curve 86 78. Equations of tangent and normal, lengths of subtangent and sub- normal. Rectangular coordinates 89 79. Parametric equations of a curve 92 80. Angle between the radius vector drawn to a point on a curve and the tangent to the curve at that point 97 81. Lengths of polar subtangent and polar subnormal 98 82. Solutions of equations having multiple roots 100 83. Applications of the derivative in mechanics. Velocity . . .102 84. Component velocities ^^^ 85. Acceleration 1^5 86. Component accelerations . . . • • • • .105 viu CONTENTS Chapter Vm SUCCESSIVE DIFFEREHTUTIOH SECTION PAOB 87. Definition of Buccessive derivatives 109 88. Notation 100 89. The nth derivative 110 90. Leibnitz's formula for the nth derivative of a product . . .111 91. Successive differentiation of implicit functions . . . .112 Chapter IX MAXIMA AHD MINIMA 92. Increasing and decreasing functions 116 93. Tests for determining when a function is increasing and when decreasing 117 94. Maximum and minimum values of a function .... 118 95. First method for examining a function for maximum and mini- mum values 121 96. Second method for examining a function for maximum and mini- mum values 123 97. General directions for solving problems involving maxima and minima 120 Chapter X POINTS OF INFLBCTION 98. Definition of points of inflection and rule for finding points of inflection 136 Chapter XI DIFFERENTULS 99. Introduction . 140 100. Definitions ....;..... 140 101. dx and dy considered as infinitesimals 141 102. Derivative of the arc in rectangular codrdinates .... 141 103. Derivative of the arc in polar codrdinates 143 104. Formulas for finding the differentials of functions . . . 144 105. Successive differentials ......... 146 CONTENTS ix Chapter XII RATES SECTION PAQB 106. The derivative considered as the ratio of two rates • • .148 Chapter XIII CHANGE OF VARIABLE 107. Interchange of dependent and independent variables . . .152 108. Change of the dependent variable 153 109. Change of the independent variable 154 110. Simultaneous change of both independent and dependent variables 156 Chapter XIV CURVATURE. RADIUS OF CURVATURE 111. Curvature '. . . 159 112. Curvature of a circle 159 113. Curvature at a point 160 114. Formulas for curvature 161 115. Radius of curvature 162 Chapter XV THEOREM OF KEAN VALUE. DTDETERMinATE FORMS 116. Rolle's Theorem 166 117. The Theorem of Mean Value 167 118. The Extended Theorem of Mean Value 168 119. Maxima and minima treated analytically 169 120. The Generalized Theorem of Mean Value 172 121. Indeterminate forms 172 122. Evaluation of a function taking on an indeterminate form . 173 123. Evaluation of the indeterminate form - 174 124. Evaluation of the indeterminate form — 177 00 125. Evaluation of the indeterminate form • oo 179 126. Evaluation of the indeterminate form oo — oo . . . .179 127. Evaluation of tJie indeterminate forms 0*, 1*, oo^ . . . 181 CONTENTS Chapteb XYI circlb of curvature. center of curvature SEOnOK PAOB 128. Circle of curvature. Center of curvature 183 129. Center of curvature the limiting position of the intersection of normals at neighboring points 186 130. Evolutes 187 131. Properties of the evolute 190 132. Involutes and their mechanical construction 191 Chapter XVn PARTIAL DIFFEREHTUTIOH 133. Continuous function of two or more independent variables . .193 134. Partial derivatives 194 135. Partial derivatives interpreted geometrically 195 . 136. Total derivatives 198 137. Total differentials 200 138. Differentiation of implicit functions . 1 . • . . 202 139. Successive partial derivatives 203 140. Order of differentiation immaterial 206 Chapter XVIII ENVELOPES 141. Family of curves. Variable parameter 208 142. Envelope of a family of curves depending on one parameter . . 208 143. The envelope touches each curve of the family at the limiting points on the curve 210 144. Parametric equations of the envelope of a family depending on one parameter . . . 211 145. The evolute of a given curve considered as the envelope of its normals 213 146. Two parameters connected by one equation of condition • .214 Chapter XIX SERIES 147. Introduction 217 148. Infinite series 218 149. Existence of a limit 220 150. Fundamental test for convergence ...'••. 221 CONTENTS xi SXCnON PAOB 151. Comparison test for convergence 222 152. Cauchy's ratio test for convergence 224 153. Alternating series . . ' 226 154. Absolute convergence 227 155. Power series 228 Chapter XX EXPANSION OF FUNCTIONS 156. Introduction 231 157. Taylor's Theorem and Taylor's Series 232 158. Maclaurin's Theorem and Maclaurin's Series .... 234 159. Computation by series 238 160. Approximate formulas derived from series 240 161. Taylor's Theorem for functions of two or more variables . . 243 162. Maxima and minima of functions of two independent variables . 246 Chapter XXI ASYMPTOTES. SINGULAR POINTS. CURVE TRACING 163. Rectilinear asymptotes 252 164. Asymptotes found by method of limiting intercepts . . . 252 165. Method for determining asymptotes to algebraic curves . . 253 166. Asymptotes in polar coordinates 257 167. Singular points 259 168. Determination of the tangent to an algebraic curve at a given point by inspection 259 169. Nodes 262 170. Cusps 263 171. Conjugate or isolated points 264 172. Transcendental singularities 265 173. Curve tracing 266 174. General directions for tracing a curve whose equation is given in rectangular coordinates 267 175. Tracing of curves given by equations in polar coordinates . . 269 Chapter XXII APPLICATIONS TO GEOMETRY OF SPACE 176. Tangent line and normal plane to a skew curve whose equations are given in parametric form 271 177. Tangent plane to a surface 273 xii CONTENTS asonoir paob 178. Normal line to a Emifaoe 275 179. Another form of the equations of the tangent line to a skew curve 277 180. Another form of the equation of the normal plane to a skew curve 277 Chapter XXm CURVES FOR REFERENCE INTEGRAL CALCULUS Chapter XXIV INTEGRATION. RULES FOR INTEGRATING STANDARD ELEMENTARY FORMS 181. Integration 287 182. Constant of integration. Indefinite integral . . . .289 183. Rules for integrating standard elementary forms .... 291 184. Trigonometric differentials 303 Chapter XXV CONSTANT OF INTEGRATION 185. Determination of the constant of integration by means of initial conditions 309 186. Geometrical signification of the constant of integration . . 309 187. Physical signification of the constant of integration . . • 310 Chapter XXVI INTEGRATION OF RATIONAL FRACTIONS 188. Introduction . 315 189. Partial fractions 315 190.. Imaginary roots 318 191. Case I 320 192. Caaell 321 193. Casein 322 194. Case IV 324 CONTENTS xiii Chaptbr xxvn niTEGRATIOH BT SUBSTITUTION OP A HEW VARIABLE. ,,_^„ RATIONALIZATIOE 195. Introduction , , 196. Differentials containing fractional powers of x only 197. Differentials containing fractional powers oi a-\- hx only 198. Differentials containing no radical except Va + ftz + x* 199. Differentials containing no radical except Vo + fti — ^ 200. Binomial differentials 201. Conditions of integrability of binomial differentials 202. Transformation of trigonometric differentials 203. MisceUaneous substitutions 329 329 330 331 332 334 334 337 339 , Chapter XXVIII INTEGRATION BT PARTS. REDUCTION FORMULAS 204. Formula for integration by parts 341 205. Reduction formulas for binomial differentials .... 344 206. Reduction formulas for trigonometric differentials . . . 349 207. To find fc^sin nxdx and (e^ cos nxdx 353 Chapter XXIX THE DEPINITE INTEGRAL 208. Differential of an area . . . ., 355 209. The definite integral 356 210. Geometrical representation of an integral 357 211. Mean value of 0(z) 358 212. Interchange of limits 358 213. Decomposition of the interval 359 214. The definite integral a function of its limits .... 359 215. Calculation of a definite integral 360 216. Infinite limits 360 217. When (x) is discontinuous 361 218. Change in limits 365 Chapter XXX INTEGRATION A PROCESS OF SUMMATION 219. Introduction 367 220. The definite integral as the limit of a sum 367 xiv CONTENTS SBOTIOK PAGE 221. Areas of plane curves. Rectangular codrdinates • • • .871 222. Area when curve is given in parametric form .... 373 223. Areas of plane curves. Polar codrdinates 876 224. Length of a curve 378 225. Lengths of plane curves. Rectangular codrdinates . . . 379 226. Lengths of plane curves. Polar codrdinates 382 227. Volumes of solids of revolution 384 228. Areas of surfaces of revolution 388 Chapter XXXI SUCCESSIVE AND PARTIAL Of TEGRATIOV 229. Successive integration 892 230. Partial integration 394 231. Definite double integral. Greometric interpretation . . . 396 232. Value of a definite double integral over a region .... 400 233. Plane area as a definite double integral. Rectangular codrdinates 402 234. Plane area as a definite double integral. Polar codrdinates . . '406 235. Moment of inertia. Rectangular codrdinates .... 407 236. Moment of inertia. Polar codrdinates 410 237. General method for finding the areas of surfaces .... 411 238. Volumes found by triple integration 415 239. Miscellaneous applications of the Integral Calculus . . .419 Chapter XXXII ORDDTART DIFFERENTIAL EQUATIOHS 240. Differential equations. Order and degree 424 241. Solutions of differential equations 425 242. Verifications of solutions 426 243. Differential equations of the first order and of the first degree . 427 244. Differential equations of the nth order and of the first degree . 435 Chapter XXXIH DTTEGRAPH. TABLE OF INTEGRALS 245. Mechanical integration 446 246. Integral curves . ' • . . 446 247. The integraph ' . . .448 248. Integrals for reference 450 INDEX 461 DIFFERENTIAL CALCULUS CHAPTER I COLLECTION OF FORMULAS 1. Formulas for reference. For the convenience of the student we give the following list of elementary formulas from Algebra, Geometry, Trigonometry, and Analytic Geometry. 1. Binomial Theorem (n being a positive integer) : I? l? lr-1 Also written : 2. nI=[n = l-2-8-4---(n-l)n. 3. In the quadratic equation ox^ + &x + c = 0, when &3 — 4 oc > 0, the roots are real and unequal ; when &^ — 4 oc = 0, the roots are real and equal ; when &3 — 4 oc < 0, the roots are imaginary. 4. When a quadratic equation is reduced to the form x> + jpz = 9, p = sum of roots with sign changed, and q = product of roots with sign changed. 6. In an arithmetical series, i = a + (n-l)d;« = ?(a + = |[2a + (n-l)(q. 6. In a geometrical series, , rl-a a(r*-l) r — 1 r — 1 1 DIFFERENTIAL CALCULUS 7. log a& = log a + log 5. 9. log ofl^n log a. 8. log - = log a ~ log 6. 10. log Va = - log a, 11. log 1=0. 12. logaa = l. 13. log- = - logo. 14. Circamference of circle = 2 srr.* 16. Volame of prism = Ba, 16. Area of circle = fcr^. 17. Volume of pyramid = i Ba. 18. Volume of right circular cylinder = icr^. 19. Lateral surface of right circular cylinder = 2 icra. 20. Total surface of right circular cylinder = 2 nrr (r + a). 21. Volume of right circular cone = ^ icr^a, 22. Lateral surface of right circular cone = ma. 23. Total surface of right circular cone = jrr (r + s), 24. Volume of sphere = | m*. 26. Surface of sphere = 4 itr^. 26. sin X = ; cos x = ; tan z = cscx secx cotx «_- ^ sinx. ^ cosx 27. tanx= ;cotx = -; cos X sm X 28. sin«x + cos'x = 1 ; 1-f tan^x = sec«x ; 1 + cot«x = C8c«x. 29. sin X = COB ( x j ; 30. sin (n — x) = sin x ; C08X = sln ^— -x^; cos (jr — x) = — cosx; tanx = cot (--«)• ^°(* - X) = - tanx. 31. sin (x + y) = Bin X cosy + cosx sin y. 32. sin (x ~ y) = sin X cos y — cosx sin y. 33. cos (x + y) = cosx cosy — sin X sin y. tan X 4- tan V ^. tanx — tan v 34. tan(x-f y)= ^*^*^ ^ . 36. tan(x-y)= ^ 1 — tan X tan y 1 + tan x tan y 2 tanx 36. sin 2x = 2 sin X cos X ; cos 2x = cos^x — sin^x ; tan 2x = l-tan^x X 2tan- XX XX 2 37. sin X = 2 sin - cos - ; cos x = cos^ — sin^ - ; tan x = 2 2' 2 2 ,^^,x 2 <* In formulas 14-26, r denotes radius, a altitude, B area of base, and s slant height. COLLECTION OF FORMULAS 3 38. co«»« = - + -coB2a5: 8m*a5 = cob2x. 2 2 2 2 39. l + co8X = 2co8>-: l-ooBX = 28m3-. 2 2 ^rt . « /I ~ COS X X /l-fcosx ^ « /r^ — C08X COBS 41. sin X + sin y = 2 sin i (2 + y) COS i (x — y). 42. sin X — sin y = 2 cos i (x + y)Bmi{x — y), 43. cosx + cosy = 2cosi(x + y)cosi(x — 2^). 44. cos X — cos y = — 2 siu i (x + y) sin i (x — y), 46. = = : Law of Sines. sin ^ sin J3 sin C 46. (j^ = W + c2 - 2 6c cos -4 ; Law of Cosines. 47. d = V(xi — x«)2 + (yi — y2)' ; distance between points (Xi, yi) and (x», yj). 48. d = — ?L±_^L±_.; distance from line -4x + By + C = to (xi, yi). 49. X = ?l^t_?? , y = Vl±yi . coordinates of middle point. 60. X = xo + x', y = yo + y' ; transforming to new origin (xo, yo)* 61. X = x' cos — y' sin 0, y = x' sin + / cos $ ; transforming to new axes mak- ing the angle $ with old. 62. X = p cos $, y = pa\n0; transforming from rectangular to polar coordinates. 63. p = VxM-T^i ^ = arc tan - ; transforming from polar to rectangular coor- dinates. ^ 64. Different forms of equation of a straight line : (a) ILzyi = yt-Vi ^ two-point form ; X — Xj Xj — Xi X t/ (h) - -f I = 1, intercept form ; a (c) y — yi = m (x — Xi), slope-point form ; (d) y ^mx-{-by slope-intercept form ; (e) X cos a + y sin a = p, normal form ; (f ) ^x + By + C7 = 0, general form. 66. tan = — ^ ~ , angle between two lines whose slopes are mi and m^. 1 + mima mi = ms when lines are parallel, and mi = when lines are perpendicular. ms 68. (x — a)^ + (y — fi)* = r^) equation of circle with center (a, /9) and radius r. 4 DIFFERENTIAL CALCULUS 2. Greek alphabet. Letters Names A a Alpha B iS Beta r 7 Gamma A S Delta E e Epsilon z ? Zeta H «7 Eta e <? Theta I ( Iota K K Kappa A X • Lambda M It Mu Letters Names N V Nu H f Xi O Omicron n IT Pi P p Rho 2 «r 9 Sigma T T Tau T V Upsilon ^ 4> Phi X X Chi •9 ^ Psi il <o Omega 3. Natural values of trigonometric functions. Angle in BadJana Angle in Degrees Sin • Cob Tan Got .0000 .0873 .1746 .2618 .3491 .4363 .5236 .0109 .6981 .7854 0« 5° lO'' 15° 20° 25'' SOP 36° 40° 45° .0000 .0872 .1736 .2588 .3420 .4226 .5000 .5736 .6428 .7071 1.0000 .9962 .9848 .9659 .9397 .9063 .8660 .8192 .7660 .7071 .0000 .0875 .1763 .2679 .3640 .4663 .6774 .7002 .8391 1.0000 GO 11.430 5.071 3.732 2.747 2.145 1.732 1.428 1.192 1.000 90° 85° 80° 75° 70° 65° 60° 55° 50° 45° 1.5708 1.4835 1.3963 1.3090 1.2217 1.1345 1.0472 .9599 .8727 .7854 Cos Sin Cot Tan Angle In Degrees Angle in Radians COLLECTION OF FORMULAS Angle in Badians Angle in Degrees Sin Cot Tan Cot Seo Cm (y> 1 00 1 00 2 9(y> 1 00 00 1 ic 180° -1 00 -1 00 Sir 2 27(y> -1 00 00 -1 lie 36(y> 1 00 1 00 Angle in Angle in Degrees Sin Cos Tan Cot Sec Csc 0« 1 00 1 00 fC 6 30«> 1 2 2 s V3 2\/8 3 2 7C 4 45« 2 V2 2 1 1 ^2 V2 3 60° V3 2 1 2 Va . V3 3 2 2V3 3 n 2 90° 1 00 00 1 DIFFEKENTIAL CALCULUS 4. Rules for signs. Quadrant Sin Cm Tan Got Seo G»e f irst . . « • + + + + + + Second. . . . + — — — — + Third .... — — + + — — Fourth. . . . — + — — + — CHAPTER II NUMBERS 5. Rational numbers. All positive and negative integers and fractions, and zero, are called rational numbers. We shall assume that the student is familiar with the most elementary properties of these numbers and their use in ordinary Arithmetical work. 6. Comparison of rational numbers with the points of a straight line. The series of rational numbers is unlimited, for between any two we can always insert as many more rational numbers as we please. Nevertheless there exist gaps everywhere in the series, as may be clearly seen if we set up a correspondence between the series of rational numbers and the points of a straight line. o a p On a straight line of indefinite length select a zero point and a definite unit of length for measuring segments. A length may then be constructed corresponding to any rational number a which we lay off to the right or left of according as a is positive or negative. In this way we obtain a definite end point P which may be considered as the point corresponding to the rational number a* We may then say, to every rational number there corresponds one and only one point on the straight line. But there are lengths which are incommensurable with a given unit of length. From Geometry we have the familiar example of the diagonal of a square whose side is the unit of length. Laying off such a length from the origin on the straight line we obtain an end point which corresponds to no rational number, f And * In aboTO flgnre a to taken as poeitiTe. t Length of diagonal of a unit square « v2. This cannot be an Integer, for no integer multi- plied by itself gires 2. Neither can it be a fraction ; for, if possible, let >^-?. (A) where a and h are integers which do not hare a common factor. Squaring both sides, a* 2-_. (B) Since a and h hare no common factor, a* and b* can hare no oonmion factor ; and (B), which says that 6> is contained twice in a*, contradicts our hypothesis (A). Therefore, since V2 !■ neither an integer nor a fraction it cannot be a rational number. 7 8 DIFFERENTIAL CALCULUS since there are infinitely many lengths which ai*e incommensurable with the unit of length, the straight line is infinitely richer in point-individuals than the series of rational numbers is in number- individuals. This comparison has led to the recognition of a certain incompleteness of the system of rational numbers, while we ascribe to the straight line completeness and absence of gaps^ that is, continuity, 7. Irrational numbers. If we wish to study the straight line arithmetically, the system of rational numbers having been found wanting, it becomes necessary to extend our system of numbers in such a way that it shall have the same completeness^ or continuity^ as the straight line. This has been done by the creation of irra- tional numbers which are defined in terms of rational numbers only. The scope of this book does not permit the development of the modem arithmetic theory of rational numbers; hence we shall only call the attention of the student to the existence of irrational numbers and to the statement : the irrational numbers completely fill up all the gaps which exist in the system of rational numbers; i.e. we assume that to every point on a straight line corresponds a number^ rational or irrational^ and conversely. Following are examples of irrational numbers : V2 = 1.4142136 ..,* logioS = 0.6989700 • • -,1 TT = 3.1415929 •., « = 2.7182814... 8. Real numbers. All rational and irrational numbers are called real numbers. These are arranged in order with respect to their magnitudes as follows: increasing as we pass from left to right. • It was shown in footnote on p. 7 that Vi cannot be a rational nomber. t Suppose this to be a rational number, then h where a and ( are positlre integers. Then a 10»«6, or 10" -6». That is, no matter what the ralues of a and 6, we would have a number whose last digit is zero equal to a number whose last digit is 5 ; this being absurd, our hypothesis that logjoS was a rational number is absurd. NUMBERS 9 The symbol > is read is greater than^ and the symbol < is read u less than. It is sometimes convenient to write a > 0, which means that a is positive ; or 6 < 0, showing that h is negative. We may also write such expressions as 3 > - 1, or - 8 < - 5, etc. The symbol > is read is greater than or equal to^ and is equiva- lent to the symbol <, read is not less than. The symbol < is read is less than or equal to^ and is equivalent to the symbol >, read is not greater than. The symbol ^ is read is greater or less than^ and is equivalent to the symbol ¥=, read is not equal to. 9. Numerical or absolute value. By the numerical value or abso- lute value of a real number we mean its value taken positively. The numerical or absolute value of a is denoted by the symbol |a|. Thus, |5| = I- 51=4-5. 10. Imaginary numbers. Consider the equation a:2 -h 1 = 0. No real number substituted for x will satisfy this equation. To overcome this difficulty, our number system must be enlarged by the creation of a new number. If i is a number such that i* = — 1, then the above equation is satisfied by substituting i or — t for X. Hence , . x=^±% is called the solution of the equation, and the new number t = V— 1 is termed the imaginary unit. If a is real, the expression a V— 1, or at, defines an imaginary number. IL Complex numbers. The sum a-\-bij where a and b are real numbers, defines a complex number. The first term belongs to the system of real numbers, while the second belongs to the system of imaginary numbers. Complex numbers suffice for all algebraic operations. 10 DIFFERENTIAL CALCULUS 12. Division by zero excluded. - is indeterminate. For, the quotient of two numbers is that number which multiplied by the divisor will give the dividend. But any number whatever multi- plied by zero gives zero, and the quotient is indeterminate ; that is, any number whatever may be considered as the quotient, a result which is of no value. - has no meaning, a being different from zero, for there exists no number such that if it be multiplied by zero the product would equal a. Therefore division by zero is not an admissible operation, 13. Only real numbers considered. Unless otherwise stated, only real numbers are considered in what follows in this book. CHAPTER III VARIABLES AND FUNCTIONS 14. Variables. A variable is a quantity to which an unlimited number of values can be assigned. Variables are denoted by the later letters of the alphabet. Thus, in the equation of a straight line, 2 + ^ = 1 a X and y may be considered as the variable coordinates of a point moving along the line. 15. Constants. A quantity whose value remains unchanged is called a aynstant. Numerical or absolute constants retain the same values in all problems, as 2, 5, Vt, tt, etc. Arbitrary constants^ or parameters^ are constants to which any one of an unlimited set of numerical values may be assigned, and they are supposed to have these assigned values throughout the investigation. They are usually denoted by the earlier letters of the alphabet. Thus, for every pair of values arbitrarily assigned to a and 5, the equation a represents some particular straight line. 16. Interval of a variable. Very often we confine ourselves to a portion only of the number system. For example, we may restrict our variable so that it shall take on only such values as lie between a and 6, where a and b may be included, or either or both excluded. We shall employ the symbol [a, 6], a being less than 6, to represent the numbers a, 6, and all the numbers between them, unless otherwise stated. This symbol [a, b] is read the interval from a to b. 11 12 DIFFERENTIAL CALCULUS 17. Continuous variation. A variable x is said to vary continu- ously through an interval [a, 6], when x starts with the value a and increases until it takes on the value b in such a manner as to assume the value of every number between a and h in the order of their magnitudes. This may be illustrated geometrically as follows : a X b O ^ P B The origin being at 0, lay off on the straight line the points A and B corresponding to the numbers a and b. Also let the point F correspond to a particular value of the variable x. Evidently the interval [a, b] is represented by the segment AB, Now as x varies continuously from a to 6 inclusive, i.e. through the interval [a, 6], the point F generates the segment AB. 18. Functions. When two variables are so related that the value of the first variable depends on the value of the second variable, then the first variable is said to be a function of the second variable. Nearly all scientific problems deal with quantities and relations of this sort, and in the experiences of everyday life we are con- tinually meeting conditions illustrating the dependence of one quantity on another. For instance, the weight a man is able to lift depends on his strength, other things being equal. Similarly, the distance a boy can run may be considered as depending on the time. Or, we may say that the area of a square is a function of the length of a side, and the volume of a sphere is a function of its diameter, 19. Independent and dependent variables. The second variable, to which values may be assigned at pleasure within limits depend- ing on the particular problem, is called the independent variable, or argument; and the first variable, whose value is determined as soon as the value of the independent variable is fixed, is called the dependent variable, or function. Frequently, when we are considering two related variables, it is in our power to fix upon whichever we please as the fwrfe- pendent variable ; but having once made the choice, no change of independent variable is allowed without certain precautions and transformations. VARIABLES AND FUNCTIONS 18 One quantity (the dependent variable) may be a function of two or more other quantities. (the independent variables, or arguments). For example, the cost of cloth is a function of both the quality and quantity ; the area of a triangle is a function of the hose and altitude; the volume of a rectangular parallelopiped is a function of its three dimensions. 20. Notation of functions. The symbol f(x) is used to denote a function of x^ and is Tesid f function of x. In order to distinguish between different functions the prefixed letter is changed, as F{x),4>{x),f(x),eUi. During any investigation the same functional symbol always indicates the same law of dependence of the function upon the variable. In the simpler cases, this law takes the form of a series of analytical operations upon that variable. Hence, in such a case, the same functional symbol will indicate the same operations or series of operations, even though applied to different quantities. Thus, if f(x) = a? - 9 z 4- 14, then /(y) = y'-9y4-14. Also f{a) = a* - 9 a 4- 14, /(6 4.1) = (6 4- 1)2 - 9(6 4- 1) + 14 = 6» - 76 4- 6, /(0)=0»-90 4-l4 = 14, /(-l) = (-l)»-9(-l)4- 14 = 24, /(3)=3*-9-3+14=-4, /(7)=7*-9-7 4-14 = 0, etc. Similarly ^(Xy y) denotes a function of x and y, and is read <t> function of x and y. If 4>(x,y) = 8in(x-{-y)y then <t> (a, b) = sin (a 4- 6), and *^ ( "o ' ) = si^i o ~ ^• Again, if F(xy y, z) = 2 a? 4- 3 y — 12 «, then F{m, — tw, w) = 2 m — 3 w — 12 m = — 13 wi, and F{S, 2, 1)= 2 3 4- 32 - 12 1 = 0. Evidently this system of notation may be extended indefinitely. 14 DIFFERENTIAL CALCULUS 21. Values of the independent variable for which a function is defined. Consider the functions a:* — 2 a: H- 5, sin a;, arc tan x of the independent variable z. Denoting the dependent variable in each case by y, we may write y = a? — 2a;4-5, y = sina:, y = arc tan x* In each case y (the value of the function) is known, or, as we say, defined^ for all values of x. This is not by any means true of all functions, as the following examples illustrating the more com- mon exceptions will show. Here the value of y (i.e. the function) is defined for all values of X except a; = 6. When a: = 6 the divisor becomes zero and the value of y cannot be computed from (1). Any value might be assigned to the function for this value of the argument. (2) y = V;. In this case the function is defined only for positive values of x. Negative values of x give imaginary values for y, and these must be excluded here where we are confining ourselves to real num- bers only. (3) y = log„a:. a > Here y is defined only for positive values of x. For negfative values of x this function does not exist (see § 85). (4) y = arc sin x^ y = arc cos x. Since sines and cosines cannot become greater than -h 1 nor less than — 1, it follows that the above functions are defined for all values of x ranging from — 1 to -f 1 inclusive, but for no other values. 22. One-valued and many-valued functions. A variable y is said to be a one-valued function of a second variable x when y has one and only one value corresponding to each value of a?. Thus, in y = 3a? y is a one-valued function of x. VARIABLES AND FUNCTIONS ' 15 If to each value of the second variable there correspond more than one value of the first variable, then the first variable is said to be a many-valued function of the second variable. In y is a two-valued function of x since Again, in y = arc tan x it is seen that there is no limit to the number of values of y corre- sponding to a given value of x. For, let a? = 0, then y = wtt, where n denotes zero or any integer. 23. Explicit functions. When a relation between x and y is given by means of an equation solved for y, then y is called an explicit function of x. Thus, in -y/x — 3 y = sin aa:, y = log(l 4- x), y = 5 a*, y is in each case an explicit function of x. Again, in 2 = log {x -f y) 2 is an explicit function of x and y. Similarly w^e'^ exhibits t£f as an explicit function of x^ y, and z. Symbolically these explicit functions may be respectively denoted z = <^(a;, y), w = F(x, y, z). 24. Inverse functions. Let y be given as a function of x by means of the relation It is usually possible in the case of functions considered in this book to solve this equation for x, giving a; = <^(y); 18 DIFFERENTIAL CALCULUS (4) Inverse trigonometric functions ; as arc sin a:,* arc cot (a; — y), etc. Many more transcendental functions are studied in the higher branches of mathematics. EXAMPLES 1. GiYen f{x) = x» - 10 x* 4- 81 x - 30 ; show that /(O) = - 30, f(y) = y» - 10y2 + Sly - 30, /(2) = 0, f(a) = a« - 10a2 -f 31 a - 30, /(3) = /(6),' f(yz) = y»«» ~ 10 yH^ + 31 yz - 30, /(I) >/(- 3), /(x - 2) = x» - 16x2 + 83x - 140, /(-I) = -6/(6). 2. If/(x) = x'-10x« + 31x-30, and (x) = X* - 66x2 ~210x- 216; showihat /(2) = 0(-2), /(3) = 0(-3), /(5) = 0(-4), /(0) + 0(0) + 246 = 0. 8. GiYen F(x) = x(x - 1) (x 4- 6) (x - J) (x + |) ; show that F(0) = F{1) = F(- 6) = F(i) = F{- i) = 0. m — 1 4. If /(wii) = — ; show that mi 4- 1 /('Wi) —f(m2) mi — mt 1 + /(wii)/(m2) 1 4- w»im« 1 -x 5. Given 0(x) = log ; show that 1 4-x 0(x) + 0(y) = 0(^^). 6. If /(0) = cos ; show that /W =/(- 0) =-/(^ - 0) = -/(«^ + 0). 7. If F(0) = tan ; show that ^ ' i-[2^(W' 8. Given ^(x) = x'" 4- x**" 4- 1 ; show that ^(1) = 3, \^(0) = 1, ^(a) = ^(-a). *A1bo written sin-ix, the -1 not being considered as a negative exponent in the ordinary sense, but merely indicating the inverse function. The expression y = arc sin x should be read y equalt the arc (or angle) whose »ine it Xf and the same relation between x and y is given by sin y^x. For example, since tan - = 1 , 4 we may also write - «> arc tan 1. 4 CHAPTER IV THEORY OF LIMITS 29. Limit of a variable. If a variable v takes on successively a series of values that approach nearer and nearer to a constant value I in such a manner that \v — l\* becomes and remains less than any assigned arbitrarily small positive quantity, then v is said to approach the limit l^ or to converge to the limit l. Symbolically this is written ^j^i^ v = L The following familiar examples illustrate what is meant. (1) As the number of sides of a regular inscribed polygon is in- definitely increased, the limit of the ai*ea of the polygon is the ai*ea of the circle. In this case the variable is always less than its limit (2) Similarly the limit of the area of the circumscribed poly- gon is also the area of the circle, but now the variable is always greater than its limit. (8) Consider the series {A) i_^ + j_j+.... The sum of any even number (2n) of the first terms of this series is 1 i 1 i i *^2ii ■■■ 2 4 8 2*""' 2**~*' ,Bs s -— -? ^ By6,p.l (20 ^..-_^_i-3 3.2..-I- Similarly the sum of any odd number (2 n + 1) of the first terms of the series is "fn+i ■*■ 2 4 8 ' 2*""' 2*"' 1 1 ((h S -_?!ll!—-2, 1 By6,p.l * To be read the numerical value qf the difference between v and L 10 20 DIFFERENTIAL CALCULUS Writing (B) and (O) in the forms ^ u « limit /2 o ^— limit ^ a we have ^ I o "" ^a« ) = q ni»^i ^ 0, and n = aoV^»-^^"3;-n = oo3T2^--®- Hence by definition of the limit of a variable it is seen that both S^^ and iS^s.^i are variables approaching § as a limit as the num- ber of terms increases without limit. Summing up the first two, three, four, etc., terms of {A)j the sums are found by (B) and ((7) to be alternately less and greater than ), illustrating the case when the variable^ in this case the sum of the terms of (il), i» sometimes less and sometimes greater than its limit. In the examples shown the variable never reaches its limit This is not by any means always the case, for from the definition of the limit of a variable it is clear that the essence of the definition is simply that the numerical value of the difference between the variable and its limit shall ultimately become and remain less than any positive number we may choose however small. (4) As an example illustrating the fact that the variable may reach its limit, consider the following. Let a series of regular polygons be inscribed in a circle, the number of sides increasing indefinitely. Choosing any one of these, construct the circum- scribed polygon whose sides touch the circle at the vertices of the inscribed polygon. Let p^ and F^ be the perimeters of the inscribed and circumscribed polygons of n sides and C the circum- ference of the circle, and suppose the values of a variable a; to be as follows : Then evidently, Tl = OO ' and the limit is reached by the variable. THEORY OF LIMITS 21 30. Infinitesimals. A variable v whose limit is zero is called an infinitenmcd.* This is written limit v = 0, and means that the successive numerical values of v ultimately become and remain less than any positive quantity however small. Such a variable is said to become indefinitely small or to idtimatelt/ vanish. If limit V = Z, then limit (v — Z) = ; that is, the difference between a variable and its limit is an infini- tesimal. Conversely, if the difference between a variable and a constant is an infinitesimal^ then the variable approaches the constant as a limit. 31. The concept of infinity (oo). If a variable v ultimately becomes and remains greater than any assigned positive number however large, we say v increases without limits and write limit t; = 4- 00. If a variable v ultimately becomes and remains algebraically less than any assigned negative number, we say v decreases without limity and write limit t; = — 00. If a variable v ultimately becomes and remains in numerical value greater than any assigned positive number however large, we say v, in numerical value^ increases without limits or v becomes infinitely great^jf and write limit v = 00. Infinity (oo) is not a number; it simply serves to characterize a particular mode of variation of a variable by virtue of which it increases or decreases without limit. * Hence a constant, no matter how small it may be, is not an InfinitMimal. t On account of the notation used and for the sake of uniformity, the expression limit v « -I- « is sometimes read v approachet the limit plua infinity. Similarly limit r » - « is read v approaehea the limit minus infinity , and limit t; » « is read v, in numericcU value, approaches the limit it^nity. While the abore notation is conrenient to use in this connection, the student must not forget that infinity is not a limit in the sense in which we defined a limit on page 19, for infinity is not a number at all. 22 DIFFERENTIAL CALCULUS 32. Limiting value of a function. Given a function f{x). If the independent variable x takes on any series of values such that limit 2; = a, and at the same time the dependent variable /(2;) takes on a series of corresponding values such that limit /(a:) = A^ then as a single statement this is written and is read the limit of f{x)^ as x approaches the limit a hy any set of values^ is A.* 33. Continuous and discontinuous functions. A function f{x) is said to be continuous for x = aii the limiting value of the function when X approaches the limit a in any manner is the value assigned to the function for 2; = a. In symbols, if then f{x) is continuou4i for x==a. The function is said to be discontinuous for x = a if this con- dition is not satisfied. For example, if limit J*/ V the function is discontinuous for x=a. The attention of the student is now called to the following cases which occur frequently. Case I. As an example illustrating a simple case of a func- tion continuous for a particular value of the variable, consider the function For a:= 1, /(a:)=/(l)= 3. Moreover, if x approaches the limit 1 in any manner, the function f(x) approaches 3 as a Umit. Hence the function is continuous for x = l. * It Bometimes happens that/Cz) approaches one limit when x approaches a, x being always less tlian a ; and a different limit when x approaches a, x being always greater than a. 0r,/(x) may approach a limit from one side and not from the other ; or it may approach no limit from either side. Evidently the above definition excludes all snch exceptional cases. THEORY OF LIMITS 23 Case II. The definition of a continuous function assumes that the function is already defined for 2; = a. If this is not the case, however, it is sometimes possible to assign such a value to the function for x = a that the condition of continuity shall be satis- fied. The following theorem covers these cases : Theorem. Iff{x) is not defined for a; = a, and \f then f(x) will be continuous for x=^a^if B is assumed as the value off(x) for x^a. Thus the function x-2 is not defined for x^2 (since then there would be division by zero). But for every other value of re, and 2; _ 2 (^ "^ ^^ ~ * * . , . limit re* — 4 therefore o K = 4. x= A X — 2 Although the function is not defined for rr = 2, if we arT)itrarily assign it the value 4 for rr = 2, it then becomes continuous for this value. A function f{x) is said to he continuous in an interval when it is continuous for all values of x in this interval,* 34. Continuity and discontinuity of functions illustrated by their graphs. (1) Consider the function rr*, and let {A) y = 2^. * In thfs book we shall deal only with f onctlons which are In general oonttnnons, that !b, con- tlnnona for all ralues of x, with the possible exception of certain isolated yalues, our resalts in general being understood as yalid only for snch ralaes of x for which the function in question is actually continuous. Unless special attention is called thereto, we shall as a rule pay no attention to the possibilities of such exceptional yalues of x for which the function is discon' tinuous. The definition of a continuous function /(x) is sometime roughly (but imperfectly) summed up in the statement that a small change in x thail produce a tmall change in /(x). We shall not consider functions baring an infinite number of oscillations in a limited region. 24 DIFFEKENTIAL CALCULUS If we assume values for x and calculate the corresponding values of y^ we can plot a series of points. Drawing a smooth line free- hand through these points a good representation of the general behavior of the function may be obtained. This picture or image of the function is called its graph. It is evidently the locus of all points satisfying equation {A). Such a series or assemblage of points is also ^ called a curve. Evidently we may assume values of X so near together as to bring the values of y (and therefore the points of the curve) as near together as we please. In other words, there are no breaks in the curve, and the function ^ is continuous for all values of x. (2) The graph of the continuous function sinx is plotted by drawing the locus of y = sin X. It is seen that no break in the curve occurs anywhere. (3) The continuous function «* quent occurrence in the Calculus. graph from y = e» we get a smooth curve as shown. ' is clearly seen that, is of very fre- If we plot its (« = 2.718 ••) From this it (a) when a;== 0, ^"^^^ y( = e^ = 1; (b) when a? > 0, y(= e') is positive and increases as we pass towards the right from the origin; (c) when x<0, y (= e*) is still positive and decreases as we pass towards the left from the origin. (4) The function log^x is closely related to the last one discussed. In fact, if we plot its graph from y = log.ar, it will be seen that its graph has the same relation to OX and OF as the graph of e* has to OF and OX. THEORY OF LIMITS 25 Here we see the following facts pictured : (a) Fora: = l, log.a; = log.l = 0. (b) For 2; > 1, log^x is positive and increases as x increases. (c) For 1 > a; > 0, log;a; is negative and increases in nwmerical value as x diminishes. (d) For 2; < 0, log^a; is not defined ; hence the entire graph lies to the right of OY. (5) Consider the function -9 and set X 1 y = — X If the graph of this function be plotted, it will be seen that as x approaches the value zero from the left (neg* atively) the points of the curve ultimately drop down an infinitely great distance, and as X approaches the value zero from the right the curve extends upward infinitely far. The curve then does not form a continuous branch from one side to the other of the axis of F, showing graphically that the function is discontinuous for 2; = 0, but continuous for all other values of x, (6) From the graph of o y = 2x it is seen that the function 2a? is discontinuous for the two values a; = ± 1, but continuous for all other values of x. (I) The graph of y = tana; shows that the function tan z is discontinuous for infinitely many values of ar, namely, a; = — , where n denotes any odd positive or negative integer. I 26 DIFFERENTIAL CALCULUS (8) The function arc tanrr has infinitely many values for a given value of Xj the graph of equation y = arc tan x consisting of infinitely many branches. If, how- ever, we confine ourselves to any single branch, ^ the function is continuous. For instance, if we say that y shall be the arc of smallest absolute value whose tangent is 2;, that is, y shall take on only values between — — and — » then we are limited to the branch passing through the origin and the condition for continuity is satisfied. (9) Similarly arc tan - X is found to be a many-valued function, branch of the graph of Confining ourselves to one y = arc tan -» X we see that as x approaches zero from the left y approaches the limit — — , and TT as X approaches zero from the right y approaches the limit + •— Hence the function is discontinuous when a; = 0. Its value for 2; = can be assigned at pleasure. Functions exist which are discontinuous for every value of the independent variable within a certain range. In the ordinary ap- plications of the Calculus, however, we deal with functions which are discontinuous (if at all) only for certain isolated values of the independent variable ; such functions are therefore in general continuous, and are the only ones considered in this book. 35. Fundamental theorems on limits. In problems involving limits the use of one or more of the following theorems is usually implied. It is assumed that the limit of each variable exists and is finite. THEORY OF LIMITS 27 Theorem I. The limit of the algebraic sum of a finite number of variables is equal to the like algebraic sum of the limits of the several variables. Theorem II. The limit of the product of a finite number of vari- ables is equal to the prodvxA of the limits of the several variables. Theorem III. The limit of the quotient of two variables is equal to the quotient of the limits of the separate variables^ provided the limit of the denominator is not zero. Before proving these theorems it is necessary to establish the following properties of infinitesimals. (1) The sum of a finite number of infinitesimals is an infinitesimal. To prove this we must show that the numerical value of this sum can be made less than any small positive quantity (as e) that may be assigned (§ 30). That this is possible is evident, for, the limit of each infinitesimal being zero, each one can be made numerically less than - (n being the number of infinitesimals) and therefore their sum can be made numerically less than e. (2) The product of a constant c and an infinitesimal is an infinir tesimal. For the numerical value of the product can always be made less than any small positive quantity (as e) by making the numerical value of the infinitesimal less thaif -• c (3) The product of any finite number of infinitesimals is an infini- tesimal. For the numerical value of the product may be made less than any small positive quantity that can be assigned. If the given product contains n factors, then since each infinitesimal may be assumed less than the nth root of e, the product can be made less than e itself. (4) If V is a variable which approaches a limit I different from zero^ then the quotient of an infinitesimal by v is also an infinitesi- mal. For if limit v = 1, and k is any number numerically less than {, then by definition of a limit, v will ultimately become and remain numerically greater than k. Hence the quotient -i where € is an infinitesimal, will ultimately become and remain numerically less than -^ and is therefore by (2) an infinitesimal. 28 DIFFERENTIAL CALCULUS Proof of Theorem L Let v^^ v^ v„ • • • be the variables, and i^^ l^ !,, • • • their respective limits. We may then write where 6^ e,, Cg, • • • are infinitesimals (i.e. variables having zero for a limit). Adding, (A) (Vj + r, + V, H ) — (^ + ^1 + ^1 H ) = («! + €t + €g H ). Since the right-hand member is an infinitesimal by (1), p. 27, we have from the converse theorem on p. 21, limit (Vj + V, + V, H ) = ?i + ?, + ?8 "• or, limit (v^ + v, + v, H ) = limit v^ + limit v, + limit v, H , which was to be proved. Proof of Theorem U. Let v^ and r, be the variables, l^ and /, their respective limits, and c^ and e, infinitesimals; then and r^ = ?, + €,. ft Multiplying, v^v, = {l^ + c^) (Z, -h €,) or. Since the right-hand member is an infinitesimal by (1) and (2), p. 27, we have as before limit (VjVj) = Z^Zj = limit Vj • limit v,, which was to be proved. Proof of Theorem lU. Using same notation as before. Vg Zg-h€, Z, \Z« + c, V or 9 THEORY OF LIMITS 29 Here again the right-hand member is an infinitesimal by (4), p. 27, if Z, =#= 0, hence limit r^V^ = ^°'^^^S \v^J i, limit V, which was to be proved. It is evident that if any of the variables be replaced by con- stants our reasoning still holds and the above theorems are true. 36. Special limiting values. The following examples are of special importance in the study of the Calculus. In the first twelve examples a> and c =#= 0. Written in the farm of limits. Abbreviated form often used. (1) (2) (8) limit £ ^ « = oo: X limit ca; = oo; c — = 00. <? . 00 = 00. limit £ _ limit ^ « = QOa. a* a' a* a* oo; -=0; (4) (5) (6) (7) (8) (9) X S'o ^°Sa a: = 4- 00, when a < 1 (10) li"^* log„ a; = - 00, when a < 1 (11 ) "^'o ^^?« a: = - 00, when a > 1 (12) x='i«^^»<»^ = + ^' whena>l 00 c c 00 — = 00. -^ = 0. limit aj =— oo limit X = + CO limit « = — « Umit a; = + CO = -f 00, when a < 1 = 0, when a < 1 = 0, when a > 1 = -f 00, when a > 1 a""" = 4-00. a+« = 0. a— = 0. a+* = -f 00, lOga = + 00. log. (+ oo) = - 00. loga = - 00. ^^ga{-h QO) = + 00. The expressions in the second column are not to be considered as expressing numerical equalities (oo not being a number) ; they are merely symbolical equations implying the relations indicated in the first column, and should be so understood. 80 DIFFERENTIAL CALCULUS limi^ !l* — fl.* (13) Find _ , where n denotes any positive integer.* By division we get ^-^ = af -* + 02^-* -h 0*2*"' -h • • • + a*"'a; + a""^ X — a for every value of x except x = a. Therefore limit ^-ar ^ limit ^-i limit ^., ^ _^ limit ^,,,^ limit ^..,^ x = a x — a x = a x = a x = a x=^a [By Theorem I, p. 27.] The limit of each term in the second member is a"~^; and since there are n terms, we have limit af - a* ^_, X = a X — a limit sin a; (14) Show that "_ ^ = L Let be the center of a circle w^hose radius is unity. Let arc-4Jlf = arc-4Jlf' = a;, and let MT and M^T be tangents drawn to the circle at M and M\ From Geometry, MPM' < MAM' < MTW\ 2 sin a; < 2 a; < 2 tan z. Dividing through by 2 sin a: we get ^ X 1 1 <-T — < sm X cos X If now X approaches the limit zero, limit X a: = si sma; lllYllf 1 must lie between the constant 1 and ^ , which is also 1. X = \) cos X rnu f ^ limit ^ i ^« limit sin a; ^ ^, ^^^ ^- Therefore ^ -: — = 1, or ^ = 1. Th. Ill, p. 27 a: = sin a; ' a; = * Restricted to a positive integer in order to simplify the work. The result holds true for all values of n. THEORY OF LlMlTS 81 It is interesting to note the behavior of this function from its graph, the locus of equation sin a; X -3«- -1 Although the function is not defined for 2; = 0, yet it is not cdscontinuous when 2; = if we define sin = 1. Case II, p. 23 (15) Find the limit of the sum of the series .11 1 1 H 1 1 1 » 3 9 3"-^ as the number of terms increases without limit. By formula 6, p. 1, we find that the sum of n terms of the series is Hence limit ^ n = 00 Umit 3 w = 00 2 ^ 3 limit A _ 1\ 2 w = 00 \ 3-y ___3r limit f't\_ limit /JLN"] ""2Ln = oo^^ w = oo V3-y J = |[l-0]-|. An. 37. The number e.* We first proceed to prove two important theorems. Theorem I. If a is a variable > — 1 which varies continuously in any interval not including zerOj then the function • <^ (a) = (1 -h a)« varies in a sense contrary to a, * The proofs in this section are due to Vall^-Poussin. 82 DIFFERENTIAL CALCULUS We start with the identity or, a"+^ = l +(a-l)(a'' + a—^4--- + a + l); where a denotes any positive number and n a positive integer. Since the last parenthesis is > or < (n + 1) according as a is > or < 1, or according as (a — 1) is positive or negative, we have in either case , ^ ^ ^ a*+^>l4-(n + l)(a~l). Let w be any number > — n and different from zero, and replace a in this inequality by the quotient 1 + - J and reducing gives ^^TTx) >^i )■ Let m be any integer > n ; then by repeated applications of the last result we see that ('^'^)' >('<)'■■ and it readily* follows that according as oi < 0. Now replacing <o by ma we get (A) a + a)l>(l+^af, according as a ^ 0. This proves our statement for two values (as a and — a) having the same sign and whose quotient is rational. In order to extend this proof to the case of two values a and /3 whose ratio is irrational, consider first the case when a > 0. * The Msumption is here made that raising both members of an inequality to any power, rational or irration€U, does not or does change the sense of the Inequality according as the power is positire or negatire THEORY OF LIMITS 83 Assuming a < /3 we may so choose m and n that n Now let — a approach yS as a limit by a series of increasing values ; then the second member of the inequality {A) will be constantly decreasing and we get in the limit since the function is continuous i i (1 -h ay > (1 -h M or, (f>{a)> <f>{/3). a</3 Similarly when a < we shall get <f>{a)<<l>(/3), a>l3 which establishes the theorem for all cases. Theorem II. Definition of the number e. As a approaches the limit zero^ the function <^(a) = (l + a)= of Theorem I approaches a limit. In whatever manner a approaches zerOy the limit is the same number. This limit is denoted by e. Consider the two variables a and /3 connected by the condition 1 + a = r -'i where > a > — 1 ; then it follows that <^(a) = (l+/8)<^(/8).* If a tends towards zero through a series of increasing (therefore negative) values, /8 will be positive and tend towards zero through a series of decreasing values. Then, by Theorem I, we know that <f>{a) continually decreases and ^(^) continually increases. But <f>(a) always remains positive and therefore must tend towards some definite limit (see Theorem II, above). Denoting this limit ' -1 • ♦(a)-(l-».a)«=(l + -i^-l)i-^^ [«-j^-lbyhypotheiiB.] 1 + 3 1 + 3 (ji-) ^ =(1 + 3) ^ -(l + 3)(l + 3)^=(l + 3)0(3). 34 DIFFERENTIAL CALCULUS by e, the last equation, in which 1 + fi tends towards the limit unity, becomes in the limit limit limit Since a was negative and /3 positive, this proves that the limit is the same whatever may be the sign of the variable. To evaluate this limit we note that ^ (ff) increases towards its limit, while <f)(a) = {l -{- fi)<f) (^) decreases towards the same limit (= e). Hence for all positive values of )S, <f>{/3)<e<{l-h^)4>{l3).^ By means of this inequality we may calculate the value of « to any desired degree of accuracy by choosing /3 sufficiently small. If we let )3 = i, then (I)' <«<(!)•; hence e certainly lies between 2 and 4. In Chapter XX, Ex. 14, p. 237, we give a more expeditious method for calculating e. Approximately ^^ ^ € = 2.71828 •••. Plotting the graph of <f>(a) from y=(l+a)% and assigning to y the value e when a = 0, we see that as a increases without limit t/ approaches the limit 1, and as a approaches the limit —1 from the right y increases without limit. Natural logarithms are those which have the number e for base. These logarithms play a very important r81e in mathematics. When the base is not indicated explicitly, the base e is always understood in what follows in this book. Thus log^v is written simply log v. Natural logarithms possess the following characteristic prop- erty: If a approaches zero as a limit in any way whatever, limit ^Qg(^"^^) ^ limit log (1 + a)* = log e = 1. ^ e ^4 1 Ss,^^^ y-e 1 y-i -1 a THEORY OF LIMITS 86 EXAMPLES Prove the following. 1. ««"'» (±±1) = 1. Proof. Un.it/x + lN umit/j 1\ limit ..V . limit /1\ m. t «^ = X = «<^>+x = ao(i) Th.I,p.27 [Since these limits exist.] = 1+0 = 1. 2 limit / g'-f 2g \ 1 Proof, »°^it /?!±2^\ ^ limi X = ap\6_3xa/ x = ( .» + - limit ( as [DiYidfiig both namerator and denominator by x>.] limit /i . ?\ x = ooV ■'"x/ limit / 6^ _ 3'\ x = «Vx2 / [Since the limit of the denominator is not zero.] limit .jv , limit /2\ X = op^ ' "*■ X = oo\x/ limit f^\_ limit .o\ X = ao\x2/ x = oo' ^ ■ [Since these limits exist.] _ 1 +0 _ 1 ~0-3" 3* Th. m, "p. 27 Th. I, p, 27 «• x = l^^Tf^ = ^- «• j[»-^J(3ax«-2Ax + 5^^) = 3ax«. 4 limit 8xM^6^ ^ _ 2 n^it , ^ 5^ ^. ,) = «. 5 limit ?1±1^5 o limit (g -k)^-2 fcc» _ ' x = -2 X4-3 • * k = x(x-\-k) »• Jj^^Q[2 8in(a + wix)co8{a-mx)] = 8in2a. ^n limit r • /tt + «\ /u-»\"| 11. "^" [t«i«(y - ») - cos V] = seca tf. If "*" " 86 DIFFERENTIAL CALCULUS . A limit cos (a — a) 1^- » — -jz ~ = - tan o. a = -cos (2 a — a) «i« limit /„^.„„ 1 +x\ tiT - . ^^ . *'^' ^ _ Q ( arc tan \ = — - , n bemg an odd integer. !*• yZ\ <*« COS Vl - yi) =^ , » being an odd integer. 16. ]'?'J I (ei + e-i) = a. 20. ]"»" ?llli = 8. Y^ limit 6x«-2z _ x = oo ^ -*• 22. ^^^^^[cos{^ + ;i)?f|^] = cos(?. 18. ^*™>t -J^ = 1. oo limit tan y = <»y + i \ ''''•0 = 19. ^^™^^ ri(n^-\) ^ ijjjj.^ 1 - cos ^ 1 n = oo(n4-2)(n + 3) ' '^^ (? = ^ — = 2* OE limit 1 .... X = a i^T^ = - 00, If X is increasing as it approaches the value o. Oft limit 1 , •, . J « = a i^TS = +«♦"« IS decreasmg as it approaches the value o. CHAPTER V DIFFERENTIATION 38. Introduction. We shall now proceed to investigate the manner in which a function changes in value as the independ- ent variable changes. The fundamental problem of the Differ- ential Calculus is to establish a measure of this change in the function with mathematical precision. It was while investigating problems of this sort, dealing with continuously varying quantities, that Newton * was led to the discovery of the fundamental prin- ciples of the Calculus, the most scientific and powerful tool of the modem mathematician. 39. Increments. The increment of a variable in changing from one numerical value to another is the difference found by sub- tracting the first value &om the second. An increment of x is denoted by the symbol Ax, read delta x. The student is warned against reading this symbol '^ delta times a:," it having no such meaning. Evidently this increment may be either positive or negative f according as the variable in changing is increasing or decreasing in value. Similarly Ay denotes an increment of y, A0 denotes an increment of ^, A/^(ar) denotes an increment of /(a;), etc. If in y =f{x) the independent variable x takes on an increment Ax, then Ay is always understood to denote the corresponding increment of the function f(x) (or dependent variable y). * Sir Isaac Newton (1642-1727), an Englishman, was a man of the most extraordinary genius. He developed the science of the Galculiis under the name of Fluxions. Although Newton had dlsooTcred and made use of the new science as early as 1670, his first published work in which it oooure Is dated 16S7, having the title Philosophiae Naturalis Principia Mathematica. This was Newton's principal work. Laplace said of it, '*It will always remain preeminent above all other productions of the human mind." t Some writers call a negative increment a decrement, 37 88 DIFFERENTIAL CALCULUS The increment Ay is always assumed to be reckoned from a defi- nite initial value of y corresponding to the arbitrarily fixed initial value of X from which the increment A2; is reckoned. For instance, consider the function Assuming 2: = 10 for the initial value of x fixes y = 100 as the initial value of y. Suppose X increases to a; = 12, that is, Ax = 2 ; then y increases to y = 144, and Ay = 44, Suppose X decreases to a: = 9, that is, Aa; = — 1 ; then y decreases to y = 81, and Ay = — 19. It may happen that as x increases y decreases, or the reverse ; in either case A2: and Ay will have opposite signs. It is also clear (as illustrated in the above example) that if is a continuous function and A2; is decreasing in numerical value, then Ay also decreases in numerical value. 40. Comparison of increments. Consider the function {A) y = a?. Assuming a fixed initial value for 2;, let x take on an increment ^x. Then y will take on a corresponding increment Ay, and we have y -f Ay = (a; -f Aa;)^ or, y -f Ay = a^ -f 2 ic • Aa; + (Aa;)*. Subtracting {A\ y =x^ (B) Ay= 2 a: • Aa; -h (Aa-)* we get the increment Ay in terms of x and Aa;. To find the ratio of the increments, divide (B) by Aa:, giving ^ = 2 a; -f Aa;. Aa; If the initial value of a; is 4, it is evident that limit ^^3 Aa; = Aa; * DIFFERENTIATION 89 Let us carefully note the behavior of the ratio of the increments of X and y as the increment of x diminishes. Initial New Increment Initial New- Increment Ay value of X value of X dx value of y value of y Ay Ax 4 6.0 1.0 16 26. 9. 9. 4 4.8 0.8 16 23.04 7.04 8.8 4 4.6 0.6 16 21.16 6.16 8.6 4 4.4 0.4 16 19.36 3.36 8.4 4 4.2 0.2 16 17.64 1.64 8.2 4 4.1 0.1 16 16.81 0.81 8.1 4 4.01 0.01 16 • 16.0801 0.0801 8.01 It is apparent that as ^x decreases Ly also diminishes, but their ratio takes on the successive values 9, 8.8, 8.6, 8.4, 8.2, 8.1, 8.01; illustrating the fact that -^ can be brought as near to 8 in value as we please by making Ax small enough. Therefore limit Ay^g Aa; = Aic 41. Derivative of a function of one variable. The fundamental definition of the Differential Calculus is : The derivative * of a function is the limit of the ratio of the incre- ment of the function to the increment of the independent variable^ when the latter increment approaches the limit zero. When the limit of this ratio exists, the function is said to be differentiable^ or to possess a derivative. The above definition may be given in a more compact form symbolically as follows: Given and assume for x some value for which f{x) is continuous. Let X take on an increment Ax ; then y takes on an increment Ay, the new value of the function being (B) y-^Ay =^f{x -h Ax). * Also called the differential coefficient or the derived function. 40 DIFFERENTIAL CALCULUS To find the increment of function, subtract (A) from (5), giving (C) Ay=/(a^-f Aa:)-/(a:). Dividing by the increment of the variable Ax^ we get ^ ' Ax Ax The limit of this ratio when Ax approaches the limit zero is, from our definition, the derivative and is denoted by the symbol -j-* dx Therefore .rrv * dy^ limit /(a? + Ax) - f(x) defines the derivative of y \prf(x)] with respect to x. From (2>) we also get dy _ limit A|^. The process of finding the derivative of a function is called differentiation. It should be carefully noted that the derivative is the limit of the ratio, not the ratio of the limits. The latter ratio would assume the form -> which is indeterminate (§ 12, p. 10). 42. Symbols for derivatives. Since Ay and Ax are always finite and have definite values, the expression Ay Ax is really a fraction. The symbol dy^ dx however, is to be regarded, not as a fraction, but as the limiting value of a fraction. In many cases it will be seen that this symbol does possess fractional properties, and later on we shall show how meanings may be attached to dy and dx, but for the present the symbol -~- is to be considered as a whole. DIFFERENTIATION 41 Since the derivative of a function of 2; is in general also a func- tion of Xy the symbol /'(x) is also used to denote the derivative of f(x). Hence, if we may write -^ =zf*(x)y ax which is read the derivative of y with respect to x equals f prime of X. The symbol dx when considered by itself is called the differentiating operator^ and indicates that any function written after it is to be differentiated with respect to x. Thus -^ or -7- y indicates the derivative of y with respect to x ; dx dx -r-fi^) indicates the derivative oif{x) with respect to x; ax -- (2 a? -f 5) indicates the derivative of 2 a? 4- 6 with respect to x. dx The symbol D^ is used by some writers instead of — • If then ax y =/(^). we may write the identities 43. Differentiable functions. From the Theory of Limits it is clear that if the derivative of a function exists for a certain value of the independent variable, the function itself must be continuous for that value of the variable. The converse, however, is not always true; functions having l)een discovered that are continuous and yet possess no derivative. But such functions do not occur often in applied mathematics, and in this hook only differentiable functions are considered, that is, func- tions that possess a derivative for all values of the independent variable save at most for isolated values. 42 DIFFERENTIAL CALCULUS 44. General rule for differentiation. From the definition of a derivative it is seen ttiat the process of differentiating a function y =^f{x) consists in taking the following distinct steps : General Rule for Differentiation First step. In the function replace x by x + Ax^ giving a new valiLe of the function^ if 4- Ay. Second step. Svhtract the given value of the function from the new value in order to find Ay (the increment of the function). Third step. Divide the remainder Ay (the increment of the func- tion) by Ax (fhe irusrement of the independent variable). Fourth step. Find the limit of this quotient^ when Ax (the incre- ment of the independent variable) approaches the limit zero. This is the derivative required. The student should become thoroughly familiar with this rule by applying the process to a large number of examples. Three such examples will now be worked out in detail. Ex. 1. Differentiate 3x< + 6. Solution. Applying the successive steps in the General Rule we get, after placing y = 3aj«+'5, First step. y + Ay = 3 (x > Ax)^ 4- 6 = 3aja + 6x-Ax + 3 (Ax)8 + 6. Cy + Ay = 3x« + 6x-Ax + J y =3x« SecoTid step, jfy + Ay = 3x« + 6x • Ax + 3(Ax)* + 6 + 6 Ay= 6x- Ax + S(Ax)*. Third step. — = 6x + 8Ax. ^ Ax Fourth st^. Jl = 6x. . Ans, ax We may also write this -^(3x« + 6) = 6x. ox Ex. 2. Differentiate x> - 2 x + 7. Solution. Place y = ^ — 2x + 7. First step. y + Ay = (x + Ax)« - 2 (x + Ax) + 7 = x» + 8x2 • Ax + 3x • (2to)2 + (Ax)« -2x-2.Ax + 7. DIFFERENTIATION 48 Second step. Third Oep, y + Ay = x« + 3x« . Ax + 3x . (Ax)« + (Ax)« - 2x y = x' - 2 X 2- Ax + 7 + 7 Ay = 8x2 . Ax + 3x • (Ax)« + (Ax)» ^ = 8x2 + 3x • Ax + (Ax)2 - 2. Ax ^\ / -2 Ax. Fourth step. '' = 8x2-2. Ara. dx «'• ^t'^- 2« + 7) = 3x2-2. Ex. 3. Differentiate — • xJ» Solution. Place c '^ X2 First «tep. c t/ ^- Az/ ~~ ■ ' ' • • ' ' (x + Ax)2 Seomd step. Ay- "" ^-" c-Ax(2x + Ax) "' (X + AX)2 X2 X2 (X + AX)2 Third step. Ay_ ^ 2x + Ax Ax~ ^ x2(x + Ax)2 Fourth step. • dy_ ^ 2x_ 2c dx~ ^ xa(x)2~ x» Arys. Or, f_r,)=-J''. 45. Applications of the derivative to Geometry. We shall now consider a theorem which is fundamental in all applications of the Differential Calculus to Geometry. Let y =/(^) be the equation of a curve AB. Consider a fixed point P whose coordinates are (a?, y). Let x take on an increment ^{=MN)\ then y takes on an increment Ay(=JBQ), the coordinates of Q being (a; 4- Aa^ y + Ay). ^ From the figure, MP = y =f(x) y and NQ^y-^Ay =-f{x 4- Ax); therefore EQ = Ay =/(a; 4- Ar) -/(a:). Draw a secant line through P and Q and a tangent line to the curve at P. Then ^ ^ FE Ax Ax = slope of secant line FQS. N X I 44 DIFFERENTIAL CALCULUS If we now let Aa: approach the limit zero, the point Q will move along the curve and approach nearer and nearer to P, the secant will turn about P and approach the tangent as a limiting position, and we may write ^ limit Ay Aa; = Aa; __ limit f(x -h Ax) —f(x) "■A2:=0 Ax '^^' tan T = ^ from (F), p. 40 = slope of tangrent line PT, Hence Theorem. The value of the derivative at any point of a curve is equal to the slope of the line drawn tangent to the curve at that point. It was this tangent problem that led Leibnitz* to the discovery of the Differential Calculus. Ex. 1. Find the slopes of the tangenU to the parabola y = sc^ at the vertex, and at the point where x = |. Sohdion, Differentiating by General Rtde, p. 42, we get dy (A) -^ = 2x = slope of tangent line at any point on curve. dx To find slope of tangent at vertex, substitute x = in (^), giving 3^ = ^• dx Therefore the tangent at vertex has the slope zero, that is, it is parallel to the axis of x and in this case coincides with it. To find slope of tangent at the. point P, where x = |, substi- tute in (^), giving dy _. dx'^' that is, the tangent at the point P makes an angle of 45^ with the axis of x, * Gottfreid Wilhelm Leibnitz (1G4&-1716) was a native of Leipzig. His remarkable abilities were shown by original investigations in several branches of learning. He was first to publish his discoveries in Calculus in a short essay appearing in the periodical Acta Eruditorum at Leipzig In 1684. It is known, however, that manuscripts on Fluxions written by Newton were already in existence, and from these some claim licibnitz got the new ideas. The decision of modern times seems to be that both Newton and lieibnitz invented the Calculus independently of each other. The notation used to-day was introduced by Leibnitz. DIFFERENTIATION 45 Use the General Rule, p. 42, in differentiating the following examples. 1. y-SxK 2. y = x2-3x. 3. y = 0x2 + ftx + c. 4. y = x*. 6. r = o^. 6. p = 2g«. 7. « =<« -2« + 3. a y = -. X 9. « =-• 10. Find the slope of the tangent to the curve y = 2 x* ~ 6 x + 5, (a) at the p^iiit where x = 1 ; (b) at the point where x = 0. Ans. (a) ; (b) — 6. 11, (a) Find the slopes of the tangents to the two curves y = 3x^— 1 and y = 2 x^ + 3 at their points of intersection, (b) At what angle do they intersect ? Ana. (a) ±12, ±8; (b) arc tan ^. An,. f = dx 6x. dy dx 2x- 3. dy dx 2ax + 6 dy dx 3x2. dr do 2 0$. dp_ dq 4q, da dt 2t- 2. dy dx 1 X2 • ds 4 dt ti' CHAPTER VI RULBS FOR DIFFERENTIATING STANDARD ELEMENTARY FORMS 46. Importance of General Rule. The Oeneral Rule for differ- entiation g^ven in the last chapter, p. 42, is fundamental, being found directly from the definition of a derivative, and it is very important that the student should be thoroughly familiar with it. However, the process of applying the rule to examples in general has been found too tedious or difficult ; consequently special rules have been derived from the 0-eneral Rule for differentiating cer- tain standard forms of frequent occurrence in order to facilitate the work. It has been found convenient to express these special rules by means of formulas, a Jist of which follows. The student should not only memorize each formula when deduced, but should be able to state the corresponding rule in words. In these formulas, u, t;, and w denote variable quantities which are functions of x, and are differentiable. FORMULAS FOR DIFFERENTIATION I ^ = 0. dx n ^ = 1. dx m —t A- — \ — — 4. — ^^!!£. dx ^ dx dx dx* --, d , . dv V —( \— —4- ^ dx "" dx dx* VI -j-iViVf'Vn) = (VtVf"Vn)'-=-^ + (VlV» • • • t?») -^ + ••• 0nX/ 0iX €miX I t V dVn + (t?iVi-..Vn-x)-=— • uX 46 RULES FOR DIFFERENTIATING 47 vn YHa Yin Yma Yin 6 n Ha la u m im nv IV IVI lYn ^""l dx dx^ ' = nx^'K d /u\ dicKv) du dv V — — u — dx dx du d /u\ dx\c) dx C dv d(CX dx\vJ dx dv dx<*^^«^) dx = iogr«6 — • v dv ^(loi,.) dx V dx^ ' «.i dv d . «v dx d . ^ = vu^'^— — I- logrti ax «te<^*"^> dv = C08V— — • dx —— (cosr) dx dv €207 I<*»»«') « dv €207 :^(cotv) dx • dv €207 ^(sect,) . dv = secvtanv—— • dx -— (cscv) ax ^ dv = — cscvcott?-— • dx dv dx 48 DIFFERENTIAL CALCULUS IVin -=— (vers v) = sin t;---* dx dx dv WW d . . . dx m -^— (arc sin t;) = dx Vl — V* dv __ d , V dx II -7-(arc cost;) = , • €iX "^1 — f;i in — (arctant;) = T -• ^ dx^ ' l + v« dv Iin T- (arc cot V) = — T-^ — ;• dv d dx Um 3=^ (arc sect?) = dx V -y/v* 1 dv d dx mV -T-(arcc8cv) = — dx V Vv* — 1 dv ^^„ d . V dx IIV -T- (arc vers v) = €to V2t; — 1;» jjYI ^ = ^. |!i, y i>eingr a function of v. dx dv dx IIYn w ~ w"' y beingr a function of x. uX €iX dy 47. Differentiation of a constant. A function that is known to have the same value for every value of the independent variable is constant, and we may denote it by As X takes on an increment Az, the function does not change in value, that is. Ay = 0, and A:r RULES FOB DIFFERENTIATING 49 But j^"'\(^)=^ = 0, I • — -o due The derivative of a constant is zero, 9 48. Differentiation of a variable with respect to itself. Let y = ^* Following O-eneral Rule^ p. 42, we have Firet step. y -f- Ay = a; -f- Aa;. Second step. Ay = Ax, Third step. t^=1- Ax FouHh step. -~ = 1- dz ' n .-. ^ = 1. dx The derivative of a variable with respect to itself ^^ unity. 48. Differentiation of a sum. Let y = w 4- 1; — w. By General Rvlej First step, y4-Ay = u4-Au4-t;4-At; — tr — Am^, Second step. Ay = Au + Av — Aw, mr. •» ^ Av Aw . Av Aw Third step, t^ = t- + -i ir" TT ^T ^ dy du , dv dw Fourth step, ^ = — - + - --. ax dx dx dx [Applying Th. I, p. 27.] in • —( -L — ._dudv dw ^ dx, '^ dac dx dx* Similarly for the algebraic sum of any finite number of functions. The derivative of the algebraic sum of a finite number of functions is equal to the same algebraic sum of their derivatives. 50 DIFFERENTIAL CALCULUS 50. Differentiation of the product of a constant and a variable. Let y=cv. By General Rule^ First step. y -f- Ay = <?(v -f- Av) = «; -f- cAv. Second step. Ay = c • At;. Third step. --^ = {? -— . Ax Ax Fourth step. -^ = <• --. dx dx [Applying Th. II, p. 27.] _-- d , . dv IT .-. ^(«') = «5^- The derivative of the product of a constant and a variable is equal to the product of the constant and the derivative of the variable. 51. Differentiation of the product of two variables. Let y = uv. By General Rule^ First step. y -f- Ay = (m -f- Au) {v -f- Av) ^uv -\-U'Av '\-V' Au 4- Au • At;. Second step. Ay^u-Av-^-v- Au -f Au • Av. mi.' J * ^y Av ^ Au ^ . Av Third step. a=^^:— 4-v-- — hAu--—. Ax Ax Ax Ax rr ^1 ^ dy dv ^ du Fourth step. ^^u-^r-^-v- — dx dx dx C Applying Tb. 11, p. 27, since when ^x approachefl zero as a limit,"! Au also approaches zero as a limit, and limit r An — ^ j = 0. I „ d , . dv . du The derivative of the product of two variables is equal to the first variable times the derivative of the second^ plus the second variable times the derivative of the first. 52. Differentiation of the product of any finite number of varia- bles. By dividing both sides of Y by uv^ it assumes the form d , . du dv — (UV) — — dx dx dx UV " u V I RULES FOR DIFFERENTIATING 61 If then we have the product of n variables we may wnte VjV, • • • V, Vj v,v, • • • v^ _dx^ d^ dx^ ' * "^ dvj rfv, dv, dv^ dx dx dx dx Multiplying both sides by v^v^ • • • v«i we get YI :r-- (ViVi . . . V») = (ViV» . • . Vn) -^ + (VlV» • • • Vn) -^ + ••• dx dx dx ax The derivative of the product of a finite number of variables is equal to the sum of all the produMs that can be formed by multiplying the derivative of each variable by all the other variables. 53. Differentiation of a variable with a constant exponent. If the n factors in YI ai*e each equal to t;, we get — (if) — dx dx — ,— . = n — • if V dx dx When v=sx this becomes dx We have so far proven VII only for the case when n is a positive integer. In § 59, however, it will be shown that this formula holds true for any value of n and we shall make use of this general result now. The derivative of a variable with a constant exponent is equal to the product of the exponent^ the variable with the exponent diminished by unity ^ and the derivative of the variable. 62 DIFFERENTIAL CALCULUS 54. Differentiation of a quotient. Let y = - > V ^ 0. V By General Rvle^ u4- Au First step. y + Ay = Second step. Ay = V -f Av u 4- At^ t^ V'Au — u-£iv v-f-Av v v{v-^Av) Third step. Fourth step. Au Av V tt — Ay ^ Aa; Ar t;(t;4-Av) du dv V u dy dx dx dx 1/^ [Applying Theorenu II and III, p. 27.] du dv V u — d /u\ doc doc Tin ... ^(»)=-^_ dx \v/ V The derivative of a fraction is equal to the denominator times the derivative of the numerator^ minus the numerator times the derivative of the denominator^ all divided by the square of the denominator. When the denominator is constant, set t; = c in Ym, giving du dx\e/ c We may also get YlUa from IV as follows: du d /u\ _ldu _dx dx\c J c dx € The derivative of the quotient of a variable by a constant is equal to the derivative of the variable divided by the constant. RULES FOR DIFFERENTIATING 63 When the numerator is constant, set i^ = c in YIU, giving dv dx\v/ V* [sine ^-1-0.] The derivative of the quotient of a constant hy a variable is equal to minus the product of the constant and the derivative of the vari- able^ divided hy the square of the variable. All explicit algebraic functions of one independent variable may be differentiated by following the rules we have deduced so far. EXAMPLES Differentiate the following. 1. y = x». Solution, ^ = -^(x^ = 3aj«. Am. by VII a dx dx ln-3.] 2. y = ax*-6aJ». Sdidion. ^ = ^((ix*-6x«) = :^(aic*)-:^(te«) by III dz dz dx dx dx dx = 4ax»-26x. Ans, by Vila 3. y = x* + 6. Solution, T^ = ~(«*) + :^(5) by III dx dx dx = ixK Ana, byVnaandl Solution. ^ = ^(3xV)-l(7ari) + -^(8x») by III dx dx dx dx =:J^x* + }x"*-f V»"*- -^'»*- by rv and Vila 6. y = (a!^-3)». Solution, ^ = 5(xa-3)<^(xa-3) by VII dx dx [i? — X* -3 and n — 6.] = 5 (x2 - 3)* . 2x = 10x(x« - 3)<. Ana, We might have expanded this function by the Binomial Theorem and then applied III, etc., but the above process is to be preferred. 64 DIFFERENTIAL CALCULUS 6. y = Vd2-aJ«. Solutwn, ^ = :^(a^-x2)* = J(a«-x«r*^(a2-x«) by VII dz ax ax [©«a^-«« and n«-J.] 7. y=(8x« + 2)VTT6i^. 5ofu«ion. ^ = (3xJ» + 2)-^(l + 6a^)* + (l + 5x2)i;^(3xa4.2) by V dx ax ax IwSx* + 2 and o » (1 + 6j:>)i] = (3x« + 2) i (1 + Sx^)-* ^ (1 + 6x2) + (l + 6x«)*6x by Vn, etc. ax = (3x» + 2) (1 + 5x«r*6x + 6x(l + 5x2)* 6x(8x» + 2),..,,r-i-r:;_46x. + i6x ^^ Vl + 5xa Vl + 6x« « a« + x« a y = Vaa-x^ (a« - x«)* -^ (aa + x«) - (a» + x«) ^ (a^ - x») * Solution. ^ = ^2 __ ^ byvm dx a2 - x2 _ 2x(o« - X2) + x(a« + x2) (a* - x2)i (If oltiplying both numerator and denominator by (o^-x*)".] (a* - xa)i Arts, 9. y = 5x* + 3x«-6. ^ = 20x« + 6x. dx 10. y = 3cxa-8dx + 6e. --^ = 6cx-8(l. dx 11. y = x«+». ^ = (a + 6)x«+»-i. dx dv 12. y = x» + nx + n. -^ = nx*-* + n. dx 13. f(x) = jx* - Jx^ + 6. /'(x) = 2x« - 3x. 14. /(x) = (a + 6)xa + ex + d. /'(x) = 2 (a + 6)x + c d 16. — (a + &x + cx«) = 6 + 2cx. dx 16. — (6ir-3y + 6) = 5my«-i-3. dy RULES FOR DIFFERENTIATING 55 17. B = «0|>0 1& V=:0o4-A 19. s^so + oot + iyt*- 20. l = l+W + c^. 21. a = 2«« + 3t + 5. 22. arrotP-W^ + c 23. r = afi. 24. r = c^ + <W» + c^. 25. y = 6«' + 4x' + 2a'. 26. y = V3« + Vx + -. 27. y = a + te + cx^ 28. y = ^ — r-^« X* x* - X - X* + a 29. y = i 30. y = (2x» + x«-5)«. 31. /(x) = (a + ea?)*. 32. /(x) = (l + 4x8)(l + 2x«). 33. /(x) = (a + X) Va-x. 34. /(x) = (a + x)'"(6 + x)-. 35. y = — X" 36. y = x(a9 + aJ2)Va«-x^. voPo do /. d8 :»0+A dl d9 :6 + 2ctf. dt :4t + 3. ds dt :3a^~2M. dr dB = 2a9. — = 3c^ + 2d« + e. dd ^ = 21x« + 10x» + 3x*. dx •' + ' dy _ dy a -^ = c — -. dx xa ^ = Jx>-6x« + 2x-* + ix-i dx ' dy _2x' + x + 2x*-8a dx" 2x* dy ^ = 6x(3x + 1) (2x« + aJ« - 5)« dx r(x) = ^(a + &x«)*. /'(x) = 4x(l + 8x + 10x«). a — 8x /'(x) = 2 Va-x /'(«) = (a + x)".(6 + x).[^ + ^] dy dx n x« + i dy a* + a«x»-4x* *>! Va2 - x^ 66 DIFFERENTIAL CALCULUS 2«* dy 8Wx»-4z» 37. y = 38. y = 39. a = 6« - »a (te (62 - x«)« a — x dy _ 2a a + z' dx" (a + «)«" ^ ds 3<« + <« (1 + «)« ctt (1 + 0« 41. f(0)=—J=- /'W = Va-W« (a - W«)* 44. «(x) = ^^^. *'(«)= '^"^ xVl + aJ» x»(l + ««)« dx L(o + x)"«(6 + x)"J (a + x)'»+i(6 + x)"+i' J* r ^g + g + ^g - g "1 _ __ g^ + aVg?"^^ ^ Vo + x - Va-x xa Va« - x« JTiiif. Rationalize the denominator tint 47. y=V2^. ^=?. dx y .Q 6^/-x 5 dy Wx a dx a?y 49. . = («.-«.).. g = -^. d0 2 V0 -, «<' + t7<' dtt v*'-* . v^-^ 61. 14 = ^ ^. — = 1 cd dv d (^ 62. p=<?±l^. dp^ jq- 2) V^TT V^-l dg (g-1)* 53. V = r — 5=7 Ll + Vl - xa J dy ____ny__ *c ~ X Vl -. aja RULES FOR DIFFERENTIATING 67 64. Given (a + «)* = a* + 6a*x + 10a«x«+ 10a^« + 6aa5* + «5 j find (o + x)* hy differeutiation. 66. ABSummg that 1 — x»+i = 1 +« + »«...+ x^, 1 — « deduce by differentiation the sum of the aeriea l + 2x + 8«' + .-. + nx»-i, n being any positive Integer. nx»+*-(n + l)x» + l , ^rri)5 • '^'^' 55. Differentiation of a function of a function. It sometimes hap- pens that y, instead of being defined directly as a function of a;, is given as a function of another variable v which is defined as a function of x. In that case y is a function of x through v and is called a function of a function. For example, if y = r r, l — ir and v = 1 — a?, then y is a function of a function. By eliminating v we may express y directly as a function of x^ but in general this is not the best plan when we wish to find -^. ax If y =/(f ), and v=: <f} (x)y then y is a function of x through v. Let x take on an increment Ax, giving V -{-Av=^<f>{x + Ax)j defining Av, and y 4- Ay =f{v -f Av), defining Ay. By multiplying both numerator and denominator of — ^ by Av we get Ay Ay At' Ax Av Ax Let Ax approach the limit zero, then Av also approaches the limit zero, and we have,* applying Th. II, p. 27, dy^dydv^ ' dx dv dx • Aflsumlng that Ao^O for Ax safficiently small bat not zero. 58 DIFFERENTIAL CALCULUS This may also be written (B) ^= At') •*'(«?). If y = /(v) and v = ^(ic), the derivative of y with respect to x equals the 'product of the derivative of y with respect to v and the derivative of v with respect to xJ* 56. Differeiitlation of Inverse functions. Let If the inverse function exists, denote it by {B) y ^f{x). Differentiating {B) with respect to y gives [AMmnlng ^(y) and/(a?) to be dlfferentlable.] -i ___dy dx ' "^ dx dy dx If then —- = 0' (y) is different from zero, we get dy dy or, 1 (i» r(x) = The derivative of the inverse function is equal to the reciprocal of the derivative of the direct function, 57. Differentiation of a*logarlthm. Let y = log^v.f Differentiating by the General Rvle^ p. 42, considering v as the independent variable, we have First step, y + ^y=: log„ (v + Av). * It is understood that y and r have fixed initial ralues corresponding to some fixed initial ▼alue of X. t The student most not forget that this function is defined only for poeitiye values of the base a and the Tarlable v. RULES FOR DIFFERENTIATING 69 Second step. Ay = log„(t; 4- Av) — log^v [By 8, p. 2.] r Dividing the logarithm by v and at the same time multiplying the exponent of thel [parentheslB by v chjuiges the form of the expression but not its value (see 9, p. 2). J du 1 Fourth step. -ii = - log„«.* Am 'when A9 approaches the limit sero, — also approaches the limit sero. Therefore ^^ v "*" — / ^^ ** *» '^^ Theorem II, p. 33, plaoing a s Hence Since v is a function of x and it is required to differentiate log^v with respect to a:, we must use formula (-4), § 56, for differentiat- ing B, function of a function^ namely, dy _ dy dv dx dv dx Substituting value of -^ from (A)^ we get dy dx = log.c 1 (2t; t; dx dv n .*. 3-(l0gaV) : dx = lOga I dx '' V ' When a = e this becomes • Ha ^(log«) = " V * * Since logflV is a continuous function. 60 DIFFERENTIAL CALCULUS The derivative of the logarithm of a variable is equal to the product of the modulus* of the system of logarithms and the derivative of the variable^ divided by the variable. 58. Differentiation of the simple exponential function. Let y^cL"' a > Taking the logarithm of both sides to the base e^ we get log y = w log a, or, v = --e^ = - logy. log a log a Differentiate with respect to y by formula Ha, ^ = JL 1. dy^loga y* and from (C), § 56, relating to inverse functions^ we get ^=loga.y, or, (.1) ^ = loga.a''. Since t; is a function of x and it is required to differentiate a" with respect to 2:, we must use formula (A)^ § 55, for differenti- ating a function of a function^ namely, dy __dy dv dx dv dx Substituting the value of -^ from (-4), we get dy y „ dv -f- = log a •a'' -5-. dx dx w <f . V - dv I ••. -T-Ctt*) = loera •«»•-—-• doD dx When a = «, this becomes la —-(«») = c*'---. daC' dx •The logarithm of « to any base a (= loggC) is called the modulus of the system whose base is a. In Algebra It is shown that we may find the logarithm of a number N to any base a by means of the formula . j^ log«iV= logac . logeA'= lofea The modulus of the common or Briggs* system with base 10 is logioe«. 434294 •". BUL£S FOR DIFFERENTIATING 61 The derivative of a constant with a variable exponent is eqtuU to the product of the natural logarithm of the constant^ the constant with the variable exponent, and the derivative of the exponent, 59. Differentiation of the general exponential function. Let y = tt'.* Taking the logarithm of both sides to the base «, log.y=vlog.tt, or, y =«"'*«•. Differentiating by formula Xa, by¥ dx , dv ] n ... A (t^f) -. ^u^-i^ + log u . i«« ^. dx dx dx The derivative of a variable with a variable exponent is equal to the sum of the two results obtained by first differentiating by YII, regarding the exponent as constant; and again differentiating by I, regarding the base as constant. Let t; = n, any constant ; then XI reduces to But this is the form differentiated in § 68, therefore YII holds true for any value of n. Ex. 1. Differentiate y = log(a5» + a). Sohitioii. dy dx dx x^ + a [r=a:«+a.] 2x = -r An». • 11 can here assume only positive ralues. bylXo 62 DIFFERENTIAL CALCULUS Ex. 2. Differentiate y = log Vl - z\ Soltrfioj*. ^ = — d* (1 _ a;!i)J bylXa _i(l -*»)"*(- 2x) (1 - x«)» by VII ■^ i^ j<4«jt aj»- 1 Ex. 3. Differentiate y = a»«*. Solution. ^^ = log o . a»«^ f (8 x«) dx dx byX = 6xloga-a««". a-m. Ex. 4. Differentiate y = 6c«"+«*. Solution. ^*^ = 6^ (c^+«^) dx dx by IV = 66^ + «*^ (c«+x«) dx by Xa = 26xe«^+«". 2ln«. Ex. 5. Differentiate y^if. SohOUm. f ^ = (FTf^- » ^ (X) + x** log x ^ Je*) dx dx d«^ by XI = e^Tf^^ + x** logx { c*j = e»*^(- + logxj. ilna. 60. Logarithmic differentlatioii. Instead of applying H and IX a at once in differentiating logarithmic functions, we may sometimes simplify the work by first making use of one of the formulas 7-10 on p. 2. Thus above Ex. 2 may be solved as follows. Ex. 1. Differentiate y = log Vl — x". Solutum. By using 10, p. 2, we may write this in a form free from radicals as follows. y = i log (1 - aJ»). Then -^ = by IX a dx 2 1-xa ^ 1 -2x X . = ::'z :: = -z • -^n*. 2 l-x2 xa-1 BULES FOR JOIFFERENTIATING 63 Ex. 2. Differentiate y 1 /nn? Schition. Simplifying by means of 10 and 8, p. 2, V = iPog(l + x«) - log(l - ai«)]. 2L l+x« l-«« J dy dx ^_ *" 2x by IX a, etc. Am. l+«» \-7fi l-«* In differentiating an exponential function, especially a variable with a variable exponent, the best plan is first to take the loga- rithm of the function and then differentiate. Thus Ex. 5, p. 62, is solved more elegantly as follows. Ex. 3. Differentiate y = x«*. Solution, Taking the logarithm of both sides, log y = e« logx. By 9, p. 2 Now differentiate both sides with respect to x. dy — = e*— (logx) + logx-^ (6«) by IXo and V y dx dx = e« — h log X • e*, or, ^ = «'-y(- + lo««) dx = cx^(- + logx). An^ Ex. 4. Differentiate y = (4 x« - 7)« + '^'^^. Solviion, Taking logarithm, logy = (2 + Vxa-6)log(4x« - 7). Differentiating both sides with respect to x, 1J? = (3 + V^36) 8^+log(4*«-7).-^ ydx 4x«-7 Vx«-6 *? _-/.-., ^v^%^Sr7r8(2 + Vx2-6) . log(4x«- dx = x(4xa-7)-^^r«^^±^^ ^n,. L 4x^-7 VSTTg J In the case of a function consisting of a number of factors it is sometimes convenient to take the logarithm before differentiating. Thus, Ex. 6. Differentiate y = ^(^-^)(^-^) . \(x-3)(x-4) Solution. Taking logarithm, logy = i[log(x - 1) + log(x - 2) - log(x - 8) - log(x - 4)]. 64 DIFFERENTIAL CALCULUS • Differentiating both sides with respect to x, 1 dy _ 1 r 1 ^l 1_ _ 1 n j 2ga-10x + ll (X - l)(x - 2) (X - 8)(x - 4)* dy 2xa-10x-U or, -=- = . An», *B (X - l)» (X - 2)* (X - 3)*(x - 4)* Differentiate the following. 1. y = log(x + a). EXAMPLES dy 2. y = log(ax + 6). 3. y = log- 1 — X 4. , l+x« 6. y = e^. dx x + a dy a 6. y = e**+ 7. y = log (x« -f X). 8. y = log(x»-2x + 6). dx ox + & dy_ dx 2 l-x* dy dx 4x 1-x* dy_ dx ac«. dy_ dx 4c<*+«. dy_ 2x + l dx x^ + x dy 3x« - 2 dx x* — 2x + 6 9. y = log. (2 « + «•). 5f = log„e|±i^. dx 2x + x> 10. y = xlogx. J? = iogx + l. dx 11. /(x) = logx». /'(x) = ?. X 12./(x) = log»x. /'(x) = ii?i!^. X Hint. log«ar«(logar)». Use tint Vn, v» logz, n= 3 ; and then IX a. 13. /(x) = log?^l^. /'(x)= ^^ a - X a* - x^ 14. /(x) = log (X + vT+x5). f (X) = -A Vl + x« BULES FOB DIFFEBENTIATING 65 16. y = c^. 16. y = b^. 17. y = 7**+«*. 18. y = c«*--^. 19. r = o«. 20. r=za^: 21. a = «*■+**. 22. u = cu/^, 23. p = e«>«««. dz = log a • o«*e*. ^ = 2«log6.6»». dy dx dy dx = 21og7(x+l)7«^+««. = — 2xlogc«c«*-«^. d$ ^ dr __ g}^ ^ log a de~ $ ^ = 2te6"+^. dt du ae^ d» 2 Vc ^ = e9i«f«(l + log5). 24. f-[e'(l-x«)] = c«(l-2« dz 26. l/glz2)= 2^ . -«»). 26. — (x«c«') = Jce<«(ax + 2). dz 27. y = log e« 1 + e« 28. y = |(cS-c »). 29. y = 30. y = z^a/'. 31. y = x*. 32. yrra^. 33. y = xJ<«*. dx~"H-c*' d2^ 4 dz (ex + c-.,a dy dz :a*x»-Mn + xl dy^ dz :X*(l0gX + l). 1 dy x*(l — logx) dz x« dy dz :10gXa.X»<«*-*. 66 DIFFERENTIAL CALCULUS 84. /(y) = logy.«i'. /(y) = ei'(logy + i) 1 36. /(x) = log(log«). /(«) = aclogx • 37. J'(*) = log«(logx). y(^)^ 41og»(logx) xlogx 88. 0(x) = log(log*«). 0'(«)= * xlogx 89. ^(y) = log.JlI|. ^'(y) = _-L_. 40. /(x) = log:^^^^l^. /'(x) = - ^ Vxa + 1 + X Vl +x« JTini. fint rationalise the denominator. 41. y = x^««*. ^ = 0. dz 42. y = e^. -?^ = €*•(!+ log x)x«. ax d© \ t> / «• «=(i)' |=(f)W«-log.-l). 47. y = x«^. --^ = x«*+"-i(nlogx + l). dx -?? = x«"x«(^logx + log«x + - V dx \ x/ 48. y = x*". 49. y = a^^5=5. ^= xylogg ^ (a« - x2)* 50. y = c*(x»-nx«-i + n(n- l)x"-* ). -?^ = c«X». dx 61 (X + 1)« tfy_ _ (x + l)(6x« + 14x + 6) (x + 2)»(x4-3)** dx (X -H 2)* (X + 3)* .ERn^ Take logarithm of both sides before differentiating in this and the following examples. RULES FOR DIFFERENTIATING 6T ICO <* - !)• 62. y = — ^2 ^ . («- 2)*(x-3)« 68. y = xVl-x(l+«). 66. y = x»(a + 3x)«(a-2»)«. dy ^ dy _ (X - l)»(7x« + 80x - 97) *» 12(x-2)*(x-8)V dy 2 + x — 5x* *» 2 Vl-x dy_ 1 + 3X^-2x* <to " (1 - x»)* = 6x* (a + 8x)«(o - 2x) (a« + 2ax - 12x«). 66. y = 2g±i!. Vx~o dy (x — 2 a) Vx + o ^ (X - a)* 61. Differentiation of sin v. Let y = sin t?. By General Rule^ p. 42, considering v as the independent variable, we have First step. y + Ay = sin (v + ^v)* Ay = sin (t; 4- Av) — sin t^ ^ / AvX . Av ♦ 8= 2 cos/ v + — J . sin — . /. Av^ Second step. Third step. Av 2 Fourth step. Slnoa and and •Let I Adding, Therefore A-¥B^2v\-^v 2 -^ = cos V. ^-o('^Vi^a4).p.3o. \ -sT / ^"« + Av B^v Suhtneting, ^ i-B^Av \{A -*-T- Sabttitnting theie valnes of ^, J7, i(^ + ^, 4(^-1?) In terms of v and Av in the formula from Itlgonometry (42, p. 8), ■in^-sln^-2coeiM + ^8ini(^-^, we get iln(» + Ar)-ilnp»2ooefr + -^8ln— -. 68 DIFFERENTIAL CALCULUS Since v is a function of x and it is required to differentiate sin t; with respect to x, we must use formula (^), § 55, for differentiating ^function of a function^ namely, dy _dy dv dx dv dx Substituting value -~ from Fourth step^ we get dy dv -f- =cos V-—- dx dx HI ••• — - (sin v) = cos V —— • iix ax The statement of the corresponding rules will now be. left to the student. 62. Differentiation of cos v. Let y = cos v. By 29, p. 2, this may be written Differentiating by formula III, dy -2. — cos dx = cos( dv = — sm V ^- dx i-")(-S) Fsinoe COB (-- v\ *» Bin v, by 29, p. 2.1 IIII ••• ^— (cos v) = — sin V— — ax ax 63. Differentiation of tan v. Let y = tan v. By 27, p. 2, this may be written sin V y = cost; I RULES FOR DIFFERENTIATING 69 Differentiating by formula vm, cos v ^- (sin v) — sin V ^- (cos v) dy dx^ ' dx^ dx cos^v ^ dv , . i dv cos* V -r- + 8in' V -r- dx dx cos* V dv dx a dv = sec V — . cos* V dx d, . dv HT .•. ^ (tan V) = 8ec«t? — • 64. Differentiation of cot v. Let y = cot V. By 27, p. 2, this may be written 1 y = T tan V Differentiating by formula YIII 6, dx tan* V IT ... ^(cot*) = -«««*»'rf^ 65. Differentiation of sect;. Let y = sec v. By 26, p. 2, this may be written 1 y = , dv sec* V -r- , dx o at^ = — csc^ V— . tan*t; aaJ dv cos V To biFFEBENtlAL CALCULUli Differentiating by formula Yin 6, dx cos^v dv sint;—- dx cos* v , dv = sec vtant;--- dx ___ ' d , dv lYI .•• :T-(8ect^) = sect; tan t7---« 66. Differentiation of esc v. Let y = CSC v. By 26, p. 2, this may be written 1 y = . sin V Differentiating by formula YUI 6, d , , . (2a; sin* v dv COS v^- _ (fa sin*v , dv = — CSC vcot V--- (fa IVn ••• -r-(C8Ct;)= — CSCVCOtV—^. ax die 67. Differentiation of vers v. Let y = vers v. By Trigonometry this may be written y = 1 — cos V. Differentiating, dy . dv -:p- = sin v — - • cea: dx ITm .% :^ (vers r) = sin v ^. dx dx RULES FOR DIFFERENTIATING 71 In the derivation of our formulas so far it has been necessary to apply the General Rule^ p. 42 (i.e. the four steps), only for the following : TTT ^ / . . du dv dw A, , . Ill -r-(M + V — f£?) = -7--|--; — 3-- Algeoraic sum. ax dx dx ax V -r-(^v) = u-7--|- V-;-- Product. dx dx dx du dv VIII ± (1\ = !5I^. Quotient. dx\vj tr dv d dx IX — (log^ v) = log.e — • Logarithm. XII -;-(sin v) = cost;-7- Sine, aa; dx XXVI ^ ^ ^ rftj Function of a function. dx dv dx Not only do all the other formulas we have deduced depend on these, but all we shall deduce hereafter depend on them as well. Hence it follows that the derivation of the fundamental formulas for differentiation involves the calculation of only two limits of any difficulty, viz., S'-T-'=i by (14), p. 30 and J^J (1 + ")* = «• By Th. II, p. 88 Differentiate the following. 1. y = sin ox^ ^ = co8ax«~(ax^ ^r XII dz ax ^ ' = 4axco«i?a^V 72 DIFFERENTIAL CALCULUS 2. y = tan Vl — x. ^ = 8ec2Vrri-^(l-x)i by XIV ox ax sec^ Vl - X 2Vl-« * 3. y = cos^ This may also J)e written y = (cos x)'. ^ = 3 (co8x)«-^ (coB«) by VII ax ax iv » COS X and na 3.] = 8 cos* X ( - sin x) by Xni = — Ssinxcofi^x. 4. y = sin nx sin" x. — = sinnx— (8inx)» + »in»x-'(8innx) by V dx dx dx ' ^ [tt— Bin nor and r » Bin" or.] (J /f = sinnaj- n(sinx)"-» — (sinx) + sin"xcosnx— (nx) by Vn and XII ax dx = nsinnx-8ln»-'xco8x + n sin" x cos nx = n sin"-! x(8in tix cos x + cos nx sin x) = n sin»- 1 X sin (nH- 1) £• e dy 5. y = secax. -^^asecoxtanox. dx 6. y = tan(ax + 6). ~ = a sec* (ox + 6). dx 7. y = sin«x. — = sin2x. dx S. y = co8»x». -^ = -6xcos*x«sina^. dx 9. f(y) = sin 2 y cosy. /'(y) = 2 cos 2y cos y - sin 2 y sin y. 10. F(x) = cot* 5 X. JP''(x) = - 10 cot 6x cosec* 6x. n. F(^) = tan^-^. F'(^ = tan9^. 12. /(0) = 0sin0 + cos0. /'(0) = 0co8 0. 13. /(Q = sin* e cos t. /' (Q = sin^ t (3 cos» e - sin* Q. 14. r = a cos 2 ^. — = — 2 a sin 2 tf. d0 RULES FOR DIFFERENTIATING 73 16. r = aVcoe2e. 16. r = a (1 — C08 ^. 17. r = a8iii«-. 3 18. —- flog cos x) = — tan X. dx 19. — Oogtanx) = -r-;--. dx 8m2x dr aain2$ d^ Vco82tf — = a sin 6, d0 dr . o^ ^ — = a sin* -COS-. d0 3 3 ^' 3-(logsln«x) = 2cotx. dx 21. y = tanx — 1 secx , /l + sin X 22. y = log-^- ; \ 1 — sin X 23. y = logtanQ + |). 24. /(x) = 8in(x + a)co8(x — a). 26. /(x) = 8in(logx). 26. /(x) = tan(logx). 27. « = cos-' 28. r = 8inl. 29. p = sin (coe q). 30. y = c^n* 31. y = o*«>"*. 32. y = e«»*8inx. dy -=^ = sin X + COS X. (ix dy 1 dx cosx dy 1 dx cosx /'(«) = cos2x. /'(3:) = cos (log x) X /'(x) = sec* (log x) X da . a asm-- dt «2 1 dr 2C08-- d& ^ -^ = — sin 9 COB (cos q), dq — = C^n*COSX. dx J^ = na*"** sec* nx log a. dx _K = 00012 ^cos X — sin* x). dx 33. y = e* log sin x. dy dx = c* (cot X + log sin x). 74 DIFFERENTIAL CALCULUS 34. JL (x«c^naf) = a?»-' e^*(ii + x cos x). dx 35. — (e««coBmx) = 6«*(aco8m» — mslnm*). dz 36. /($) = 1 + CO8 1 — COB0 2 sin 9 ^,^v _ c°»(a Bin »- cos ») 38. /(«) = (« cot «)». 39. r = |tanS0-tAn0 + tf. 40. y = a^«. 41. y = (sin x)*, 42. y = (sinx)««*. 43. y = x + logcosTx - ^V J v/ — (1 - cos 6)^ /'W = :€9^Bin^. /'W = 28cots(cot« — a cosec* a). tan«tf. dy^ dx . ^/sinx . , X«*n*( 1- log \ X XCOBxY dy -^ = (sin x)* [log sin X + X cot x]. dx - - = (sin x)*^*(l + sec* x log sin x). dx dy 2 dx 1 + tanx 44. From sin 2 x = 2 sin x cos x, deduce by differentiation cos 2 X = cos* X — sin* x. . n+1 , nx sin— -—xsin — 2 2 45. From sin x + sin 2 x + • • • + sin nx = :: ♦ deduce by differentiation cosx + 2cos2x+"* +n cos nx = sin- n+1 , X . 2n+l sin - sm x 2 2 2 -K sin n + 1 )* sin*- 2 [n«- a poBitiTe integer.] 68. Differentiation of arc sin v. Let then y = arc sin v ; * V = sin y. * It Bhould be remembered that this function is defined only for raluef of v between - 1 and + 1 IncluBive and tliat p (the function) ii many-Talued, tliere being infinitely many arcs whose sines all equal v. Thus, in the figure (the locus of y =s arc sin r), when r = OAf, y - AfPj, Af/*„ Af/>„ • • •, Af ^j, MQ^, ••-. In the above discussion, in order to make the function slngle-ralued, only yaluee of y between -' and ' indusiye (points on arc QOP) are considered; that ISj^the arc z ^ of smallest numerical value whose sine is v^ kULES ^OR btFFfe&ENtlATti^Gt Yfi Differentiating with respect to y by in, dv — = cosy; dy dy_ 1 dv cos y But since v is a function of x^ this may be substituted in therefore By (C), p. 68 giving dy __ dy dv dx dv dx dy 1 dv da; cosy dx 1 rfr (-4), p. 57 -yflZr^dx [000 y-" VI -Bin* y - v^l-r»; the positlye sign of the radical being taken*1 since cosy is positive for all values of y between -^ and ^ inclusive. I HI d dv doc ^ (arc sin v) = - ^ doc Vl — v« 69. Differentiation of arc cos v. Let y = arccost;;* then t; = cos y. Differentiating with respect to y by nn, dv dy di^ 1 = - sin y ; therefore dv siny By (C), p. 68 - Y But since t; is a function of 2:, this may be substi- tuted in the formula dy __ dy dv dx dv dx* {A), p. 57 * This function is defined only for values of r between -1 and + 1 inclusive, and is many-valued. Ii^the figure (the locus of y»aroco8v), when vOM, In order to make the function single-valued, only values of y between and ir inclusive are considered ; that is, the smallest positive arc whose cosine is V. Hence we oonfine ourselves to arc QP of the graph. 76 DIFFERENTIAL CALCULUS giving dy 1 dv dx siny dx dv Vl-v*^ XX [Blny-%^l-ooi*y=v^l-i>*, the pluB sign of the radical being takenl Blnoe ainy ia poaitive for all yaluea of y between and v inoli]alYe.J dv d , doc ••• -— (arc cos v) = • doc Vl — V* 70. Differentiation of arc tan v. Let y = arctant>;* then V = tan y. Differentiating with respect to y by XIY, dv , — = sec"y; dy therefore dy^ 1 dv sec* y By (C), p. 68 But since v is a function of x^ this may be substituted in the formula giving dy dy dv dy 1 dv (2i; sec'y (ir 1 dv \-\''^ dx [aec* y = 1 -I- tan* y « 1 + «•.] (^), P- 57 XXI d_ dK (arctant;) = dv dx l+r« • This function ia defined for all yaluea of v and is many- valued, aa is clearly shown by its graph. In order to make it single-yalned, only yaluea of y between- \ and ^ are conaid- ered ; that ia, the arc of smalleat numerical yalue whose tangent lav (branch ^02))* RULES FOR DLFFERENTIATING 77 7L Differentiation of arc cot t;.* Following the method of last section, we get dv xin <i . .V doc •-r-(arc cot V) = — -r-- — -. dao 1 + 1>* 72. Differentiation of arc sec v. Let y = arc sec v ; t then t; = sec y. Differentiating with respect to y by ITI, dv — = secytany; therefore dv sec y tan y By (C), p. 68 But since v is a function of sc^ this may be substituted in the formula dy dy dv (A), p. 57 * Thii fnnetion to defined for all yalues of v and 1b many-Talued, aa la leen from Its graph (Fig. a). In order to make ft slngle-yalned, only raluee of y between and ir are ooniidered ; that is, the smallest positive arc whose cotangent is v. Hence we confine ourselyes to branch AB, Flo. 6 t This function is defined for all yalnes of v except those lying between -1 and + 1, and Is seen to be many-valued. To make the function slngle-ralued, y is taken as the arc of smallest numerical value whose secant Is v. This means that if v is positive we confine ourselves to points on arc AB (Fig. b), y taking on values between and | (0 may be Included) ; and if o is negative we confine ourselves to points on arc DC^ y taking on values between -v and ~- (-« may be included). 78 DIFFERENTIAL CALCULUS giving av xim dx sec y tan y dx 1 dv t; Vv^— 1 rf^ riecy-r,Midtaiiy- v^8eo«y-l=^r»-l, the pins gig;]! of the"! I radical being taken Binoe tan p 1b poBltiye for all yalnea of y I [between and ~ and between - v and - ^ , including and - v. I dv d , . dx die t, Vv« — 1 • • 73. Differentiation of arc esc v.* Following method of last section, my d dv dx — - (arc CSC v) = — . dx t?Vv* — 1 74. Differentiation of arc vers v* Let y = arc vers t;;t . then V = vers y. • Thla function 1b defined for all yaluee of v except those lying between -1 and +1, and ia seen to be many-valued. To make the function single-yalued, y is taken as the arc of Bmalleat numerical ralue whose secant is v. This meana that if v is posltire we confine ourselres to points on arc AB (Fig. a), y taking on values beween and ^/f may be included\ ; and if v is 2\2 / , negative we confine ounelves to points on arc CD, y taking on values between -v and -- ( - ^ may be indudedV 2 Fio. a FIO. ft t Defined only for values of v between and 2 induBlve, and is many-valued. To make the function continuous, y is taken as the smallest positive arc whose versed sine is v ; that is, y lies between and v inclusive. Hence we confine ourselves to arc OP of the graph (Fig. 6). RVLES FOR DIFFERENTIATING 79 Differentiating with respect to y by XYm, dv therefore f = ^' By (C), p. 58 But since v is a function of Xj this may be substituted in the formula dy 1 dv giymff -^ = dx Biny dx 1 dv ^2v — v^^^ tfbi y * Vl-coB*tf " ^l-(l-veray)« = v^2 1? - r», the plus sign of the radical beingl taken Blnoe Biny la positive for all values of y between and v Inclusive. J dv d , ^ doc HV .•. -— - (arc vers v) = doc V2t; — r* Differentiate the following. 1. y = arctanax>. SolvJUm. cgf^ «* by XXI (iy_ i^'^ dx 1 + (ox*)* = 2ax l + a^x* -4»«) • dy__ ^(8x 4«.) dx vr: (3x- 4x8)3 [» -3a;-4a<.] , 3-12x« 2. v = aicain(3x — 4x'). A>iii£i(m. ?= byxix 3 Vl - 0x2 + 24x*- 16x0 Vl-x^ 80 DIFFERENTIAL CALCULUS x^ + l 3. y = arc sec dx\za-l/ Solution. ^ = ^ ;'" "^ by XXm xa_i\\xa-l/ (x«--l)2z-(a;a + l)2g (xa - 1)« 2 4. y = arc sin - x^+j. 2x x« + 1 x2-l*xa-l dy 1 6. y = arc cot (x* — 5). a dz Va« - x* dy — 2x 6. y = arc tan dz 1 + (x* - 6)a 2x dy 2 7. y = arc cosec 1 - x^ dx 1+ x« 1 dy 2 8. y = arcyer82x'. 2x«-l dx Vl-x? dy 2 9. y = arc tan Vl — x. dx Vl - x« dy 1 <^ 2 Vl-x(2-x) lA 8 dy 2 10. y = arc cosec — " 11. y = arc vers 2x dx V9-4x* 2x3 dy 2 1 + x« dx 1 + xa io * 35 dy a 12. y = arc tan - * " 13. y = arc sin a dx a^ + z^ . x + 1 dy_ 1 V2 d* Vl -2x-x« 14. /(x) = X Va2 - x3 + a^arcsin -. /'(x) = 2 Va« -x^. a a — x\l 16. /(x) = Va* - x2 + aarcsin -. /(x) = ("^^^^^ a \a + x/ 16. X = r arc vers?- — V2 ry — y*. dx 17. tf = arc8in(8r-l). »• dy V2ry-y* d^ 3 dr V6r-9r* 18. = arc tan 19. 8 = arc sec RULES FOR DIFFERENTIATING 81 r + g d»_ 1 1 -or' dr ""l + f«* 1 da 1 Vl -t* <tt Vi-t« d X 20. -—(xarcsinx) = arc8iIlx + <*3t Vl-x* ^ d o tan tf 21. — (tan $ arc tan 0) = sec* ^ arc tan -\ — - da^ ' 1+^ d 1 22. —[log (arc cos i)] = dt arc cos t Vi — f-* 23. /(y) = arc cos (logy). f(y) = - y Vl - (log y)a 24. /(d) = arc sin Vsind. /(«) = ^ vTTcosecS. + CO8 26. p = C^*"9. 27. u = arc tan 28. 8 = arc cos dq l + g« e" — e-» du 2 2 du C + r-T c« - e-* d« 2 e' + c-' dt c< + e-' .^ dy . /arcsmx . logx \ dx \ X Vi - x*'^ 30. y = c**arctanx. -J^^c^f:; - + x* arc tan x(H- logx) |. dx Ll + x* J 31. y = arc sin (sin x). T" = ^• dx ^4 sin X 32. y = arc tan 3 + 6 cos X a , jx — a 33. y = arccot- + log\/ \x + a 34. X lfX + l+x\l 1 36. y = Vl - x' arc sin x — x. x«»~l 36. y = arc cos dy 4 dx 5 + 3 cos X dy 2ax* dz X* — a* dy _ X2 dz 1-x* dy _ X arc sin x dx Vl-x2 dy 2nx"-i aj2" + 1 dx x2" + 1 82 DIFFERENTIAL CALCULUS Formulas {A), p. 57, for differentiating 9^ function of a function^ and (C), p. 58, for differentiating inverse fanctionSj have been added to the list of formulas at the beginning of this chapter as XXVI and XXVII respectively. In the next eight examples, first find -^ and —- by differentiation and then uv dx substitute the results in *? = ^.* by XXVI dv • dz dv dx to find -^ . dx In general our results should be expressed explicitly in terms of the independent variable ; that is, — in terms of x, — in terms of y, --^ in terms of 0, etc. dx dy d$ 37. y = 2t>«-4, c = 3x« + l. — = 4«; ~ = 6x; substituting in XXVI, dv dx ^ = 4c.6x = 24x(8x« + l). dx 38. y = tan 2 v, v = arc tan (2 x — 1). — = 2seca2t; — = ^ ; substituting in XXVL dv 'dx 2x2-2x + l ® ^ dy_ 2sec«2p tan'2i?-fl _ 2xg-2x-fl (ix""2x*-2x + l ~ 2x2-2x + l~ 2(x-x2)2 Since va arc tan (227-1), tan»=2j;-l, tan2r= 'I 39. y = 3t>«-4« + 6, t = 2x»-5. ^ = 72x6 -204x». dx -^ 2d X dy 4: 8«-2 2x-l dx (x-2)« dy 41. y = log (a^ — «^, v = asinx. ^ = ^2 tan x. 42. y = arc tan (a + «),« = e^. dx dy e* dx 1 + (a + c»)* 43. r = ^» + c», « = log(«-««). ^ = 4«»-6ea + l. 44. to = 4 log -^ ^ -arc tan -— » y/l + Sz-\-Sz^ dw 1 t) = dz «c (1 + z) * As was pointed out on p. 67, it might be possible to eliminate v between the two given expressions so as to find y directly as a function of jr, but in most cases the above method is to be preferred. RULES FOR DIFFERENTIATING 8S dx In the following examples first find — by differentiation and then snhstitute iu dy dz"^ dz to find ^^. dz dy 46. x = yVl + y. 46. x=vrT cosy. dy 2 VI +y _ 2x dz 2 + Sy 2| r + 3y* 2 dy_ 2 Vl + cos y siny dz %^ jP« 47. x = -JL_. ^^ (l + logv)« 1 + log y dx log y ^o 1 « + Va» — y" dy y 48. X = a log — . i' _ » dx ^Qi^^ 49. x = rare vers?- V2;:ir^. *? = ^/?IEi. r dx \ y 60. r = — ^ gr^^t^-^). 1 + logA d« «> Ki 1 «* + Ve««'-4 61. u = log — ! . * 2 62. Show that the geometrical significance of XXVII is that the tangent makes complementary angles with the two codrdinate axes. 75. Implicit functions. When a relation between x and y is given by means of an equation not solved for y, then y is called an implicit function of x. For example, the equation ar»-4y=:0 defines y as an implicit function of x. Evidently x is also defined by means of this equation as an implicit function of y. Similarly x' + f + z'-.a!' = defines any one of the three variables as an implicit function of the other two. It is sometimes possible to solve the equation defining an implicit function for one of the variables and thus change it into an explicit function. For instance, the above two implicit functions may be solved for y giving or y=4 and y=±Va^-ar^-«*; 84 DIFFERENTIAL CALCULUS the first showing y as an explicit function of x^ and the second as an explicit function of x and z. In a given case, however, such a solu- tion may be either impossible or too complicated for convenient use. The two implicit functions used in this article for illustration may be respectively denoted by and F{^^ y, z) — 0. 76. Differentiation of implicit functions. When y is defined as an implicit function of x by means of an equation in the form it was explained in the last section how it might be inconvenient to solve for y in terms of x\ that is, to find y as an explicit function of X so that the formulas we have deduced in this chapter may be ap- plied directly. Such, for instance, would be the case for the equation {B) aa* -f 2 ^y-y'x-l^ = 0. We then follow the rule : IHfferentiatey regarding y as a function of x^ and put the result equal to zero.* That is, ' {€) ^/(x, y) = O. Let us apply this rule in finding -^ from {B). ^Jaa? + 22^y-y''x-l(i) = 0; by (C) dx dx (2 2^ - 7 x/)^ = y^ - 6 arr^ - 6 2;^y ; dx Ans, dy ^y"^ — % aal^ — 6 u^y dx~ 2 7?^lxf The student should observe that in general the result will con- tain both x and y, * Tills procem will be justified in % 138, p. 202. Only corresponding values of x and y which satisfy the given equation may he substituted In the derivative. KULES FOR DIFFERENTIATING 85 EXAMPLES Differentiate the following by the above rule. dy 2p 1. y^ = ^px. 2. za + y2 = T«. 3. 62a;a4.aV = a2 6*. 4. y8-3y + 2ax=:0. 5. x^ + y* = a*. a y«-2a;y + 6« = 0. 9. x« + y' — 3 oxy = 0. 10. X9z=y^. 11. p2 = a«C082^. 12. pScos0=:a38in39. 13. COS (uv) = cv. 14. = cob(0 + 0). dx V dy_ dx X — — • V dy dx b»x a^y dy 2a dx 3(1 -y«) dy dx -4 dx ^x dx a^ dy dx V V- X dy ay- -x« dx ya- - ox dy y^- - ay log y dx xa- - xy log X d^ a^ sin 2 P dp 8a9 cos 3 ^ + p8 sin tf de 2pCOB0 du c + 11 sin (uv) dv — 1 D sin (uv) de sin (9 + 0) d0 1 + sin (0 + ^) CHAPTER VII SIMPLE APPLICATIONS OF THE DERIYATIVE 77. Direction of a cttnre. It was shown in § 45, p. 44, that if y =/(^) is the equation of a curve (see figure), then dy dx = tan T = slope of line tangrent to curve at any point P. The direction of a curve at any point is defined to be the same as the direction of the line tangent to the curve at that point. From this it follows at once that dy dx = tan T = slope of curve at any point P. At a particular point whose coordinates are known we write slope of curve (or tangent) at point (xi^ f^i). y=Vx At points such as 2>, Fy JJ, where the curve (or tangent) is parallel to the axis of JT, T = 0% therefore ^ = O. ax At points such as Ay B, Gj where the curve (or tangent) is perpendicular to the axis of X^ T = 90% therefore ^ = «. ax 86 SIMPLE APPLICATIONS OF THE DERIVATIVE 87 At points such as E^ where the curve is rising^* a positive number. T = an acute ansrie, therefore -^ = ax The curve (or tangent) has a positive slope to the left of B^ between D and Fy and to the right of G. At points such as (7, where the curve isfallinffy* T = an obtuse ansrle, therefore —^ = a negative number. The curve (or tangent) has a negative slope between B and 2> and between F and G, Ex. 1. Given the curve y = — — x* + 2 (see flgore). 3 (a) Find t when x = 1. (b) Find r when x = S, (c) Find the points where the curve is parallel to OX. (d) Find the points where r = 45°. (e) Find the points where the curve is parallel to the line 2x-3y = 6 (line -4^. dy Solution. Differentiating, — = 2b> — 2 x = slope at any point. dx (a)tonT=r^l = 1 -2 = -1; therefore T = 186° Ans. (b) tan T=-~ =9 — 6 = 8: therefore t = arc tan 8. Ans, Ldxj Xa 3 (c) T = 0°, tan T = — = ; therefore x" — 2 x = 0. Solving this equation, we And dx that X = or 2, giving points C and D where curve (or tangent) is parallel to OX. (d) T = 45°, tan T = — = 1: therefore x« - 2x = 1. Solving, we get x = 1 ± V2, dx giving two points where the slope of curve (or tangent) is unity. (e) Slope of line = | ; therefore x* — 2 x = f . Solving, we get x = 1 ± Vj, giving points E and F where curve (or tangent) is parallel to line AB. Since a curve at any point has the same direction as its tangent at that point, the angle between two curves at a common point will be the angle between their tangents at that point. Ex. 2. Find the angle of intersection of the circles {A) x2 + ya-4x = l, (B) X» + y*-2y = 9. * When moving from left to right on carve. 88 DIFFERENTIAL CALCULUS Solutu and (1, - y m. Solving simultaneously we find the points of intersection to be (3, 2) -2). dx y r ^^ **"= * from(B). By § 76, p. 84 W^ "2 = — i = slope of tangent to y Jx=8 ^'2 (A) at (3, 2). j:> ^rr^ r_ 1 = — S = nInnA nf tAncrAnt: to ITC\ at l^ 2^ |f=2 The formula for finding the angle between two lines whose slopes are mi and m^ is fill — Wlj tan^ = « 1 + w,\in% — i 4- 3 Substituting, tan ^ = — UL- = i ; therefore 6 = 45°. Aia, This is also the angle of intersection at point (1, — 2). 65, p. 3 EXAMPLES The corresponding figure should be drawn in each of the following examples. 1. Find the slope oiy — 1 + xa at the origin. A%i», 1 = tan r. 2. What angle does the tangent to the curve x^^ = a^ (x + y) at the origin make with the axis of X? Aw&. r ^ 135°. 3. What is the direction in which the point generating the graph of y = 3 x^ — x tends to move at the instant when x = 1 ? ^n«. Parallel to a line whose slope is 6. 4. Show that — (or slope) is constant for a straight line. dx 5. Find the points where the curve y = x' — 3x* — 9x + 5 is parallel to the axis of X. Am. X = 3, X = — 1. 6. Fiud the poijits where the curve y (x — 1) (x — 2) = x — 3 is parallel to the axis of X, Aivi, x = 3 ± V^. 7. At what point on j/^ = 2 x^ is the slope equal to 3 ? Am. (2,4). 8. At what points on the circle x^ + y' = r^ is the slope of tangent line equal to -f ? A-M ■ ( 3r 5 / 9. Where is the tangent to the parabola y = x>-7x + 3 parallel to the line y = 6x + 2? Aim. (6, -3). 10. Find the points where the tangent to the circle x^ + y^ = 169 is perpen- dicular to the line 5x + 12 y = 60. Am. (±12, T 5). SIMPLE APPLICATIONS OF THE DERIVATIVE 89 11. Find the point where the tangent to the parabola ^ = 4 ox is parallel to the Jine x-\- y = 2, ' Ans, (o, — 2 a). 12. At what angles does the line 3 y - 2 x — 8 = cut the parabola y^ = Sx. Am. arc tan \^ and arc tan \, 13. Find the angle of intersection between the parabola ^^ = 6 x and the circle x2 + y2 _ 16, ^^^ arc tan f Vs. X^ «/2 14. Show that the hyperbola x^ -- y* = 5 and the ellipse f- — = 1 intersect at right angles. x' 15. Show that the circle x' + ^ = 8 ox and the cissoid v' = 2a — X (a) are perpendicular at the origin ; (b) intersect at an angle of 45° at two other points. 16. Find the angle of intersection of the parabola x^ = 4 ay and the witch 8a' y = Am. arc tan 3 = 71° 33'. 9. x* + 4 a* 17. Show that the tangents to the folium of Descartes x' + y^ = 3 axy at the points where it meets the parabola y^ = ox are parallel to the axis of F. 18. At how many points can the curve y = x' — 2x2 + x — 4 be parallel to the axis of X? What are the points ? An». Two ; at (1, - 4) and (\, - V^). . 19. Find the angle at which the parabolas y = 3 x^ — 1 and y = 2 x^ + 3 intersect. Am. arc tan ^. 20. Find the relation between the coefScients of the conies aix' + 5iy^ = 1 and Oax* + ha/^ = 1 when they intersect at right angles. . 1_1_1^1 ai 6i Oa 6i 78. Equations of tangent and normal, lengths of subtangent and subnormal. Rectangular coordinates. The equation of a straight line passing through the point (a^^, y^ and having the slope m is y-y^^m{x- x^. 64 (c), p. 3 If this line is tangent to the curve AB at the point Pi{x^y y^), then from § 77, p. 86, _dy* = tanT = r$?l =1^^ \_dxj :r= xi dx^ M N X m \__dxj V-Vl Hence at point of contact Py (orj, y^ the equation of the tangent line TP^ is • By this notation is meant that w© should first find -^, then in the result substitute x-^ for x (Ix J and w, for y. The student is warned against interpreting the symbol — to mean the (ierira- rt.r, tive o/ffi teieh respect to x^ for that has no meaning whatever since Xj and f/i are both constants. 90 DIFFERENTIAL CALCULUS The normal being perpendicular to tangent, its slope is -i=-S- By65,p.3 And since it also passes through the point of contact Fy^{x^y y^), we have for the equation M N X of the normal P^N That portion of the tangent which is intercepted between the point of contact and OX is called the length of the tangent (= TP^^ and its projection on the axis of X is called the length of the subtangent (= TM). Similarly we have the length of the normal (= P^N) and the length of the subnormal (= MN), MP , In triangle TP^My tan t = -— i ; therefore MP dxi (8) TM* = = yi -^ = lengrth of subtangrent. ^ ' tan T dyi MN In the triangle MP^N^ tan t = -rr^; therefore (4) MN\ = MP^ tan T = yi ^ = lengrth of subnormal. The length of tangent (== TP^ and the length of normal (= P^N) may then be found directly from the figure, each being the hypote- nuse of a right triangle having the two legs known. Thus (5) = y 1 -%/ ^ — i ^ + 1 = lengrth of tangrent. P,N = y/MP^ + & = ^(y,)» + (y, ^ J (6) = yi -Jl + ( ^ )* = lengrth of normaL The student is advised to get the lengths of the tangent and of the normal directly from the figure rather than by using (5) and (6) as formulas. * If Bubtangent extends to the right of T, we consider it positire ; if to the left, negative, t If subnormal extends to the right of My we consider It positlre ; if to the left, negative. SIMPLE APPLICATIONS OF THE DERIVATIVE 91 EXAMPLES 1. Pind the equations of tangent and normal, lengths of subtangent, sabnonnal tangent, and normal at the pomt (a, a) on the cissoid jfi = 2a — z Solution. dy __ 3 ax^ — afi dx~ y(2a-x)^ dj/i rdyl 3 a* — a" Hence ^> = ^ = _3^i!ji:*L = 2 = dope. dzi LdxJx^a a(2a-a)« Sabstituting in (1) gives y = 2x — a, equation of tangent. Substituting in (2) gives 2 y + X = 3 a, equation of normal. Substituting in (3) gives d TM = - = length of subtangent. Substituting in (4) gives MN = 2 a = length of subnormal. Also, FT = V( TMf + (ilfP)2 = -y/— + a^ = ? Vs = length of tangent, and PN = V(3fiV^)2 4. (MP)^ = vToM^ = a Vs = length of normal. 2. Find equations of tangent and normal to the ellipse x^ -^ 2y^ — 2xy — z = at the points v^here x = 1. Ara. At (1, 0), 2 y = x — 1, y + 2 x = 2. At (1, 1), 2y = x+ 1, y + 2x = 3. 3. Find equations of the tangent and normal, lengths of subtangent and sub- normal at the point (Xi, yi) on the circle x^ + y^ = r^. ^ Ans. xix 4- yiy = r^, Xiy — yjx = 0, — — 1 Xi. Xi 4. Show that the subtangent to the parabola y' = 4 px is bisected at the vertex, and that the subnormal is constant and equal to 2 p. 5. Find the equation of tangent at (xi, yi) (a) to the ellipse ^ + ^- =1 ; (b) to the hyperbola — - ^ = 1. a* 63 63 6. Find equations of tangent and normal to the witch y = point where x = 2 a. 8a« at the 4 o^ -f x2 Ans, x + 2y = 4a, y = 2x--3a. _ at __» 7. Prove that at any point on the catenary y =. -{(^ ■\- e *") the lengths o\ subnormal and normal are - (6 " — e <> ) and ^ respectively. 4 a 92 DIFFERENTIAL CALCULUS 8. Find the equation of tangent to the conic ox^ + 2bxy + cy* + 2(Ix-f2ey +/= at the point (xi, yi). Ans. axix + b (yix + xiy) 4- cyiy + d(xi + x) + e{yi + y) +/= 0.» 9. Show that the equation of tangent to cunre (-j +(^) =2 at the point (a, &) is - + - = 2 for all values of n. a b 10. Prove that the length of subtangent toy = O'Ib constant and equal to log a 11. Get the equation of tangent to the parabola y^ = 20x which makes an angle of 46° with the axis of x. Ans, y = x + 5. Hint. FInt find point of contact by method of Ex. 1 (e), p. 87. 12. Find equations of tangents to the circle x^ + ^ = 62 which are parallel to the'line2x + 3y = 6. Ara. 2x + 3y±26 = 0. 13. Find equations of tangents to the hyperbola 4x9 — 9 y* + 36 = which are perpendicular to the line 22^ + 6x = 10. Ans. 2X'- 6y ±S = 0. i4. Show that in the equilateral hyperbola 2 xy = a^ the area of the triangle formed by a tangent and the codrdinate axes is constant and equal to a^. 15. Find equations of tangents and normals to the curve y^ = 2x^ — x* at the points where x = 1. Ans. At (1, 1), 2 y = x + 1, y 4- 2 x = 3. At (1, -1), 2y = -x-l, y-2x = -3. 16. Show that the sum of the intercepts of the tangent to the parabola X* -f y* = a* on the cobrdinate axes is constant and equal to a. 17. Find the equation of tangent to the curve x* (x + y) = a* (x — y) at the origin. Ans. y = X. 18. Show that for the hypocycloid x' + y' = a' that portion of the tangent included between the coordinate axes is constant and equal to a. X 19. Show that the curve y = ae^ has a constant subtangent. 20. Show that the length of tangent is constant in the tractrix X = vc2 - ya + - log ^ 2 c+V^^^^ 79. Parametric equations of a cunre. Let the equation of a curve be (A) F{x,y)=0. If X is given as a function of a third variable, a say, called a parameter^ then by virtue of (A) y is also a function of a, and * In Ex8. 3, 6, and 8 the student should notice that if we drop the tubecrlpts in equations of tangents they reduce to the equations of the curves themselves. SIMPLE APPLICATIONS OF THE DERIVATIVE 93 (C) the same functional relation (A) between x and y may generally be expressed by means of equations in the form (x =/(a), (^ \y=<t>(a}i each value of a giving a value of x and a value of y. Equations (B) are called parametric eqiuitions of the curve. If we eliminate a between equations (-B), it is evident that the relation (A) must result. For example, take equation of circle a? H- y* = r*, or y = Vr' — 2?, Let 2;=rcosa; then y = r sin a, and we have 'x=r cos a, j/=r sin a, as parametric equations of the circle in figure, « being the parameter. If we eliminate a between equations (C) by squaring and adding the results, we have 2^ -{-if^ =z r^(cos^ a -f- sin' a) = r^, the rectangular equation of the circle. It is evident that if a varies from to 27r, the point -P(a-, y) will describe a complete circumference. In § 84, p. 104, we shall discuss the motion of a point P, which motion is defined by equations such as \y = (^(«). We call these the parametric equations of the path, the time t being the parameter. Thus in Ex. 2, p. 106, we see that Xz=Vq cos a • ^, l/ = -i^fft^ -\-v^sin a-t are really the parametric equations of the trajectory of a projectile, the time t being the parameter. The elimination of t gives the rectangular equation of the trajectory y = a: tan a — ff^ 2 Vq cos'^ a 94 DIFFERENTIAL CALCULUS Since from (B) y is given as a function of a^ and a as a function of iT, we have dy dy da dx da dx by XXVI da dx by XXVII da that is, (1» dy dy _ da dx " doc da Hence, if parametric equations of a curve are given, we can find equations of tangent and normal, lengths of subtangent and sub- normal at a given point on the curve, by first finding the value of -~ at that point from (I>) and then substituting in formulas ClX (1), (a), (8), (4) of last section. Ex. 1. Find equations of tangent and normal, lengths of subtangent and sub- normal to the ellipse (E) at the point where 4> = - 4 !x = a cos 0, y = 6 sin 0,* dz Solution. The parameter being 0, -- = — a sin 0, -=^ = 6 cos 0. cUp a0 Substituting in (D), -=^ = :— 5 = slope at any point. dx asin <f> *A8 in figure draw the major and minor anziliary droles of tlie ellipse. Throagh two points B and C on the same radius draw lines parallel to the axes of codrdinates. These lines will intersect in a point P (x, y) on the ellipse, because a:= OA = OB cos ^ = a cos ^ and y ^ AP= OD=OC Bin it^b^n^f Y ^ 3 /^y^"^^^ /4 11 ^ J yj or, cr t/ -=GOs6 and -=sinA. a ^ b Now squaring and adding we get — + ?^=coB«0 + Bin«<^=l, a* b* the rectangular equation of the ellipse. ^ Is sometimes oalled the eccentric angle of the ellipse. SIMPLE APPLICATIONS OF THE DERIVATIVE 96 as the point of T . Sabstituting = - iu the given equations {E)^ we get 4 contact Hence Sabstituting in (1), p. 80, (v^' V2) dVi dxi h a y- V2 =-^(*-^> or. Substituting in (2), p. 00, or. bx + a]^ = V2 a&, equation of tangent ^ _ ? / _ q \ V2 (ox — 6y) = a" — 62, equation of normal. Substituting in (3) and (4), p. 00, — ( — ^ = ^ = length of subnormal. '2 V a/ aV2 62 V2 b ^ = = length of subtangent _ M V2 Ex. 2. Given equation of the cycloid* in parametric form ix = a(5 — sin 6), y = a (1 — cos 6) ; 6 being the variable parameter. Find lengths of subtangent, subnormal, tangent, and normal at the point where ^ = - • 2 Solution, dx dy — = a(l — cos ^, -^ = o sin ^. de ^ ' de *The path described by<a point on the droomference of a circle which rolli without sliding on a fixed straight line is called the cycloid. Let the radius of rolling circle be a, /> the gener- ating point, and M the point of contact with the fixed line OJT, which is called the base. If arc PM equals OM in length, then P will touch at O if circle is rolled to the left. We hare, denoting angle PCM by 0, a: = OJf - J^^= o« - a sin « = o(« - sin tf ) , y= P^= AfC- -4C=a- acos tf=a(l- cos ») ; the parametric equations of the cycloid, the angle B through which the rolling circle turns being the parameter. 0D^2 ira is called the base of one arch of the cycloid, and the point V is called the yertez. Eliminating 9, we get the rectangular equation x^aarcoosT — ^j- V2ay->y>. 96 DIFFERENTIAL CALCULUS Substituting in (D), p. 94, ~ = = slope at any point. dx 1 — cos d Since 5 = - 1 the point of contact is ( a, a V and -— = 1. 2 \ 2 / dxi Substituting in (3), (4), (6), (6) of last section, we get length of subtangent = a, length of subnormal = a, length of tangent = a \^, length of normal = a V^. Aim. Note. Draw the tangent PT, the vertical diameter MB^ and connect P and B. ^ , , ^ 2 sin - cos - ^ X ^rn.r^ ^V sin ^ 2 2 tan MTF = -^ = = — = cot -• dx 1-cos^ ^ . ^0 2 2 sin^ - [From 37, p. 2, and 39, p. 3.] Hence angle MTF = ^ - h ' By 29, p. 2 Q Ako, angle PBM= -, since it is measured by one half the arc MP which measures the central angle 0^ and we have ansfle APB = — — - . ^ 2 2 Comparing, we see that angle MTP = angle APB. Therefore : The tangent to a cycloid always passes through the highest point of the generating circle. EXAMPLES In the follo'wing curves find lengths of (a) subtangent, (b) subnormal, (c) tangent, (d) normal, at any point. ^ rm. {x = a (cos ^ + ^ sin t), 1. The curve \ , • , ^ / Ana, (a) ycoitj (b) ytant, sin t cos t SIMPLE APPLICATIONS OF THE DERIVATIVE 97 2. The hypocycloid (astroid) X = 4 a 006* t, y = ia sin' t Ana, (a) — y cot t, (b) — y tan t. sin t cos t « r»«- J- .J (« = a(2co8t - C08 2Q, 3. Thecardioid { /o • * • o/ ( y = a (2 sin t — sin 2 1). 80. Angle between the radius vector drawm to a point on a curve and the tangent to the curve at that point. Let the equation of curve in polar coordinates be p =/(^). Let F be any fixed point (/>, 0) on the cui-ve. If 0y which we assume as tbe independent variable, takes on an increment Ad, then p will take on a corresponding increment A/). Denote by Q the point (p + A/), d^ A0). Draw PE perpendicular to OQ. Then OQ^zp-^^Ap^ o PR = p sin Ad, and OR = p cos Ad. Also, . _,__ PR PR p sin Ad tan POR = = = ::• ^ RQ OQ-OR /) -f- A/3 - /3 cos Ad Denote by -^ the angle between the radius vector OP and the tangent PT, If we now let Ad approach the limit zero, then (a) the point Q will approach indefinitely near P; (b) the secant PQ will approach the tangent PT as a limiting posi- tion; and {c) the angle PQR will approach ylr as a limit. Ad = /) -f A/) — /» COS Ad ^ limit P sin Ad Ad = _ . ,Ad ^ zpsur— -f A/» j since from 39, p. 3, p-p oos A*- p (1 - OM A*)- 2p lin*-^..] ^ limit Ag A^ = 0„ . ,A^ ^^^ Ap [Diyiding both numerator and denominator by A0.] 98 DIFFERENTIAL CALCULUS limit P- sin Ad . Ad . Ad '"'T Ap c;„«^ limit /^P\ ''P „„a limit / •„ ^^\ a i ^^"^'^ Ad = o(^Ai; = i ^"'^ Ad = o("°TJ = ®' *^ limit / sin Ag \ Le = (S\ A0 J (-4) . A0 sin-— = land^li=^\ L: A^ 2 = 1 by (14), p. 30, we have From the triangle OFT we get (JB) r = e + tlr. Ex. 1. Find ^ and r in the cardioid p = a (1 — cob 9). dp Solvtion, -^ = a sin 9. Substituting in (^) gives 2 a sin2 - p a(l-cos^) 2 tan^ = -^ = — ^^ = ^^^— ^-^— dp aame ^ . dS 2 a sin - cos 2 2 = tan -. By 89, p. 3, and 37, p. 2 6 0^0 Since tanf = tan-i ^ = -. Ana. Substituting in (B), t = ^ + - = — Ans, A Jt 2 2 81. Lengths of polar subtangent and polar sub- normal. Draw a line NT through the origin pei^ pendicular to the radius vector of the point P on the curve. If FT is the tangent and FN the normal to the curve at P, then and of the curve at P. OT = length of polar subtangent^ 0N= length of polar subnormal SIMPLE APPLICATIONS OF THE DERIVATIVE 99 OT In the triangle OPT^ tan -^ = Therefore P (7) or = /> tan-^ = /)*— = len^h of polar subtangent.* up In the triangle OFN^ tan -^ = -^. Therefore (8) ON^ — ^-- = — ^ = lencrth of polar subnormaL tan'^ »^ The length of the polar tangent (= FT) and the length of the polar normal (= PN) may be found from the figure, each being the hypotenuse of a right triangle. Ex. 1. Find lengths of polar subtangent and sabnormal to the lemniscate f^ = a^ cos 2 e. Solution, Differentiating as an implicit function with respect to 6, 2p^ = -2a»sin2(^, or*? = -?^^l^^ '^de de p Snhstitating in (7) and (8), we get p* length of polar subtangent = ^ 1 * *^ ' * a«sin2^ length of polar sabnormal = • P If we wish to express the results in terms of 6^ find p in terms of 6 from the given equation and substitute. Thus , in ab ove, p = ± a Vcos 2 ; therefore length of polar subtangent = ± a cot 2 ^ Vcos 2 6, EXAMPLES 1. In the circle p = r sin 9, find rp and r in terms of 0. Ana. ^ = ^, r = 2 ^. 2. In the parabola p = a sec^ -i show that r + ^ = ir. 3. Shpw that ^ is constant in the logarithmic spiral p = e^'. - Since the tangent makes a constant angle with the radius vector this curve is also called the equiangular spiral. $ 4. Given the conchoid p =:a sin* - ; prove that r = 4 ^. o 6. Show that tan ^ = ^ in the spiral of Archimedes p = a0. Find values of \f/ when^ = 2ir and4ir. Ans, f = 80° 57' and 85*» 27'. • When » increaaes with p, — is positive and i^ is an acnte angle, as lu above figure. Then the dp subtangent OTls positive and is measured to the right of an observer placed at O and looking along OP. When — is negative the subtangent is negative and is measured to the left of the observer. 100 DIFFERENTIAL CALCULUS 6. Find ^ in the curves p* = a» sin nd and p» = 6" cos n$, Ans. rp =:n0 and „ + n^. 7. Show tliat the curves in the preceding example intersect at right angles. Hint, Find r for each curve and compare. 8. Prove that the spiral of Archimedes p = a0t and the reciprocal spiral p = - , intersect at right angles. 9. Find the angle between the parabola p = a sec' - and the straight line p sin ^ = 2 a. Ara. 45°. 10. Show that the two cardioids p = a (1 + cos ^) and p = a (1 — cos 0) cut each other perpendicularly. 11. Find lengths of subtangent, subnormal, tangent, and normal of the spiral of Archimedes p = a^. ^ u* P^ * P ^nr, — ? '^ Ans. subt. = — , tan. = - va^ + p2, a a subn. = a, nor. = Vcfi~+~^, The student should note the fact that the subnormal is constant. 12. Get lengths of subtangent, subnormal, tangent, and normal in the logarithmic spiral p=a9. p I j Ana. subt. = ——, tan. = p-\/l + ; — r— » log a \ log2 a subn. = p log a, nor. ='p Vl -f log* a. When a = 6 we notice that subt = subn., and tan. = nor. 13. Find the angles between the curves p = a(l + cos 5), p = 6(1 — cos $). Ana. and - • 2 14. Show that the reciprocal spiral p = - has a constant subtangent. 15. Prove that the curves p» = a* co8(nd — a) and p^ =ia* cos {n0 — /3) Intersect at an angle a — j9. 16. Show that the area of the circumscribed square about the cardloid p = a(l — cos^) formed by tangents inclined 46° to the axis is f { (2 + Vs)a^. i 82. Solution of equations having multiple roots. Any root which occurs more than once in an equation is called a multiple root. Thus 3, 3, 3, — 2 are the roots of (A) a:* - 7 a:* + 9 a:* + 27 a: - 54 = ; hence 3 is a multiple root occurring three times. Evidently (A) may also be written in the form (a; - 3)« (a: 4- 2) = 0. SIMPLE APPLICATIONS OF THE DERIVATIVE 101 Let f{x) denote an integral rational function of x having a mul- tiple root a, and suppose it occurs m times. Then we may write {B) f{x)=={x^ar<f>{x), where <l>{x) is the product of the factors corresponding to all the roots of f{x) differing from a. Differentiating (5), / (x) = (a; - a)-* <^' (x) + <^ (2:) m (rr - a)^-\ or, (C) f{x) = {x- a)-> [{X - a) 4>' (X) + <^ (x) m]. Therefore /'(a:) contains the factors (x — a) repeated w — 1 times and no more ; that is, the highest common factor (H.C.F.) of f{x) and f (x) has w — 1 roots equal to a. In case f{x) has a second multiple root fi occurring r times, it is evident that the H.C.F. would also contain the factor (a: — ^8)'""'*, and so on for any number of different multiple roots, each occur- ring once more in /(a:) than in the H.C.F. We may then state a rule for finding the multiple roots of an equation f{x) = as follows : First step. Find f (x). Second step. Find the K C.F, off(x) and f {x). Third step. Find the roots of the H.CF. Each different root of the H.C.F. will occur once more inf{x) than it does in the H.C.F. If it turns out that the H.C.F. does not involve x^ then/(a:) has no multiple roots and the above process is of no assistance in the solution of the equation, but it may be of interest to know that the equation has no equaU i-e. multiple^ roots. Ex. 1. Solve the equation «» - 8x* + 13x -6 = 0. Solulum. Place f{z) = x* - 8 «» + 13 x - 6. First step. f(z) = 3 x^ ~ 16 x + 13. Second step. H.C.F. = x — 1. Third step. x - 1 = 0. .-. x = 1. Since 1 occurs once as a root in the H.C.F. it will occur twice in the given equa- tion ; that is, (x - 1)' will occur there as a factor. Dividing x" — 8x* + ISx — 6 by (x — 1)2 gives the only remaining factor (x — 6), yielding y the root 6. The roots of our equation are then 1, 1, 6. Drawing the graph of the function, we see that at the double root X = 1 the graph touches OX but does not cross it.* *Sinee the flrat deriyative yanishefl for every multiple root, it follows that the axis of X is tangent to the graph at all points corre- sponding to multiple roots. If a multiple root occurs an even number of times, the graph will not cross the axis of X at such a point (see figure) ; If it occurs an odd number of times, the graph will cross. 102 DIFFERENTIAL CALCULUS EXAMPLES Solve the first ten equations by the method of this section. 1. x>-7x« + 16x-12 = 0. Am. 2,2,3. 2. z*-6x2-8x-3 = 0. Am, -1,-1,-1,8. 3. x*-7x« + 9x2 + 27x-54 = 0. Aub. 3, 3, 3, - 2. 4. x*-6x«-9x« + Six -108 = 0. Ans, 3,3,3,-4. 6. x* + 0x" + x2-24x + 16 = 0. Am. 1,1,-4,-4. 6. X* - 9x» + 23x2 -3x- 36 = 0. Am. 3,3,-1,4. 7. x*-6x» + 10x2-8 = 0. Am. 2, 2, 1±V3. 8. x»-x*-6x« + x2 + 8x + 4 = 0. Am. -1,-1,-1,2,2. 9. x* - 16x« + 10x2 + 60x - 72 = 0. Am. 2, 2, 2, - 3, - 3. 10. x6 - 3x* - 5x» + 13x2 + 24X + 10 = 0. Am. - 1, - 1, -1, 3 ± VTl. Show that the following four equations have no multiple (equal) roots. 11. x» + 9x2 + 2x- 48 = 0. 12. X* - 15x2 -10x + 24 = 0. 13. X* - 3x« - 6x2 + i4x + 12 = 0. 14. x»-a» = 0. 16. Show that the condition that the equation x» + 3g'x + r = shall have a double root is 4 g* 4- r2 = 0. 16. Show that the condition that the equation x> + 3i)x2 + r = shall have a double root is r (4p8 -^ r) = o. 83. Applications of the derivative in mechanics. Velocity. Con- sider the motion of a point P describing a ""^ curve AB, Let « be the distance measured y along its path from some fixed point as A to any position of P, and let t be the corre- — X sponding elapsed time. To each value of t corresponds a position of P in the path and therefore a distance (or space) «. Hence 8 will be a function of <, and we may write SIMPLE APPLICATIONS OF THE DERIVATIVE 103 Now let t take on an increment A^ ; then. a takes on an increment A«, and {A) — = magnitude of the average velocity * of P during the time interval At. If F moves with uniform motion, the above ratio will have the same value for every interval of time and is the speed (= magnitude of the velocity) at any instant. For the general case of any kind of motion, uniform or not, we define the speed v (= magnitude of the velocity) at any instant as A« the limit of the ratio — - as At approaches the limit zero ; that is. At .. limit A« _ v= A^ A "r-» or, Ae = At The speed (= magnitude of the velocity) for afiy motion is the deriv- (Uive of the distance (= space) with respect to the time. To show that this agrees with the conception we already have of speed, let us find the speed (= magnitude of the velocity) of a falling body at the end of two seconds. By experiment it has been found that a body falling freely from rest in a vacuum near the eailh's surface f oUows approximately the law {B) s = 16.1 e, where s = space fallen in feet, t = time in seconds. Apply the General Rule^ p. 42, to {B). First step. «4.A«= 16.1 (t + At)* = 16.1 t»-f. 32.2 <• At +16.1 (Ae)». Second step. A« = 32.2 1 . At + 16.1 (At)^ A8 Third step. — = 32.2 1 -f- 16.1 At = average speed (= magnitude of the average velocity) throughout the time interval At reckoned from any fixed instant of time.^ Placing t = 2, A« {C) — = 64.4 4- 16.1 At = average speed throughout the time interval At aft^r two seconds of falling, * Velocity ia defined as the time rate of change of place, and is a rector quantity, t £a being the space or distance passed over In the time A<. 104 DIFFERENTIAL CALCULUS Our notion of speed tells us at once that (C) does not give us the actual speed at the end of two seconds ; for even if we take At very small, say ^^^ or j-^-(^ of a second, (C) still gives only the average speed during the corresponding small interval of time. But what we do mean by the speed at the end of two seconds is the limit of the average speed when At diminishes towards zero; that is, the speed at the end of two seconds is, from (C), 64.4 ft. per second. Thus even the everyday notion of speed which we get from experi- ence involves the idea of a limit, or in our notation V = ^^ ( A^ ) ~ ^^'^ ^** P®^ second. The above example illustrates well the notion of a limiting value. The student should be impressed with the idea that a limiting value 13 a definite^ fixed value, not something that is only approximated. Observe that it does not make any difference how small 16.1 A^ may be taken ; it is only the limiting value of 64.4 4- 16.1 At, when A^ diminishes towards zero, that is of importance, and that value is exactly 64.4. 84. Component velocities. The coordinates x and ^ of a point P moving in the XY plane are also functions of the time, and the motion may be defined by means of two equations, Tjl^^^- These are the parametric equations of Z/^^^ ^^ V^^ (see § 79, p. 92). ■-"" f I The horizontal component v^ of t; f is the — ^ i velocity along OX of the projection J!f of P, ^' and is therefore the time rate of change of X. Hence, from (9), p. 103, when s is replaced by x, we get (10) «^* = ^- at In the same way we get the vertical component, or time rate of change of y, (11) ..=f. * The equation of the path in rectangular coiirdinates may be found by eliminating t between these equations. t The direction of t; Is along the tangent to the path. SIMPLE APPLICATIONS OF THE DERIVATIVE 105 Representing the velocity and its components by vectors, we have at once from the figure ««> "=g=V(f)'+(f)"' giving the speed (= magnitude of velocity) at any instant. If T be the angle which the direction of the velocity makes with the axis of X^ we have from the figure, using (9), (10), (11), dy day dy (18) sinr = -^ = -7-; cost = — = -—; tanr = -^ = -t— V d« V dS^ Vao rf» dt dt dt 85. Acceleration. In general v wil^ be a functipn of t and we Now let t take on an increment A^, then v takes on an increment Av, and Av — = magnitude of the average acceleration* of P during the time interval Af. For any kind of motion we define the magnitude of the accelera- Av tion a at any instant as the limit of the ratio — as A^ approaches the limit zero ; that is. __ limit /AvN Ae = o\^Ae/ ^ = Ae -'--»'^^' dv The magnitude of the acceleration for any motion is the derivative of the velocity with respect to the time. 86. Component accelerations. Following the same plan used in finding the component velocities, we get for the component accelera- tions parallel to OX and OF, ,^-v dVac dVu A-i giving the magnitude of the acceleration. * Acceleration is defined as the time rate of change of Telocity, and is a vector quantity. 106 DIFFERENTIAL CALCULUS EXAUPLSS 1. By experiment it has been found that a body falling freely from rest in a vacuum near the earth's surface follows approximately the law d = 16.1<2, where s = space (height) in feet, t = time in seconds. Find magnitudes of the velocity and acceleration (a) at any instant ; (b) at end of the first second ; (c) at end of the fifth second. Solution, {A) «=16.ie>. ds (a) DifferentiaUng, — = 32.2 1, or, from (9), dt {B) V = 82.2 1 ft. per sec. /?« Differentiating again, — = 32.2, or, from (14), (O o ?= 32.2 ft. per (sec.)«, which tells us that the acceleration of a falling body is constant ; in other words, the velocity increases 82.2 ft. per sec. every second it keeps on falling. (b) To find V and a at the end of the first second, substitute t = 1 in (B) and (C) ; V = 32.2 ft. per sec., a =r 82.2 ft. per (sec.)^. (e) To find v and a at the end of the fifth second, substitute t = 6 in (B) and (C) ; ' v = 161 ft. per sec, a = 32.2 ft. per (sec.)«. 2. Neglecting the resistance of the air, the equations of motion for a projectile are X = Vi cos <f> • t, y = Vism<p-t — 16.1fi; ^^^ ^>^ where Vi = initial velocity, <p = angle of projection with ^ ^v horizon, t = time of fiight in seconds, x and y being meas- A X ^^Li^d in feet. Find the magnitudes of velocity, accelera- tion, component velocities, component accelerations (a) at any instant ; (b) at the end of the first second, having given Vi = 100 ft. per sec., = 80° ; (c) find dii-ection of motion at the end of the first second."" Solution. From (10) and (11), (a) Vx = Vi cos 4>; t>y = ©i sin — 32.2 1 Also, from (12), v = Vrja _ 64.4 1 Vi ain + 1036.8 «». From (16) and (16), oa- = ; Oy = - 82.2 ; o = - 32.2. (b) Substituting < = 1, uj = 100, = 30° in these results, we get t>x = 86.6 ft. per sec. ax = 0. Vf, = 17.8 ft. per sec. oy = - 82.2 ft. per (sec.)«. V = 88.4 ft. per sec. o = - 32.2 ft. per (sec.)«. u 17 ft (c) T = arc tan -^ = arc tan — ^ = 11°36'.9 = angle of direc- tion of motion with the horizontal. SIMPLE APPLICATIONS OF THE DERIVATIVE 107 3. If a projectile be given an initial velocity of 200 ft. per sec. in a direction inclined 45° with the horizontal, find (a) the magnitude of the velocity and direction of motion at the end of the third and sixth seconds ; (b) the component velocities at the same instants. Conditions are the same as for Ex. 2. Ans, (a) When t = 3, t> = 148.3 ft per sec, t= 17^85', when t = 6, t> = 160.5 ft. per sec., t = 159® 53'; (b) when t = 8, r, = 141.4 ft. per sec., Vp = 44.8 ft. per sec. when i = 6, «« = 141.4 ft. per sec., ©^ = — 51.8 ft. per sec. 4. The height (= s) in feet reached in t seconds by a body projected vertically upwards with a velocity of Vi ft. per sec. is given by the formula 8 = tJi«-16.1t«. Find (a) velocity and acceleration at any instant; and, if ui = 300 ft per sec., find velocity and acceleration (b) at end of 2 seconds ; (c) at end of 16 seconds. Resistance of air is neglected. Ans. (a) « = ©i — 32.2 «, o = — 32.2 ; (b) V = 235.0 ft. per sec. upwards, a = 32.2 ft per (sec.)* downwards ; (c) V = 183 ft per sec. downwards, a = 32.2 ft. per (sec.)* downwards. 6. A cannon ball is fired vertically upwards with a muzzle velocity of 644 ft per sec. Find (a) iU velocity at the end of 10 seconds ; (b) for how long it will continue to rise. Conditions same as for Ex. 4. Ans, (a) 322 ft. per sec. upwards ; (b) 20 seconds. 6. A train left a station and in t hours was at a distance (space) of « = «» + 2«« + 3t miles from the starting point. Find its acceleration* (a) at the end of t hours; (b) at the end of 2 hours. Ans. (a) o = 6< + 4 ; (b) a = 10 miles per (hour)*. 7. In t hours a train had reached a point at the distance of it* — 4t* + 10<» miles from the starting point (a) Find its velocity and acceleration, (b) When will the train stop to change the direction of its motion ? (c) Describe the motion during the first 10 hours. ^ ^ ^ ^ , .^ ^ Ana. (a) t> = t« - 12t* + 32t, o = 3<* - 24< + 32 ; (b) at end of fourth and eighth hours ; (c) forward first 4 hours, backward the next 4 hours, forward again after 8 hours. 8. The space in feet described in t seconds by a point is expressed by the formula « = 48«-16t*. Find the velocity and acceleration at the end of li seconds. Ans. r = 0, o = - 32 ft. per (sec)*. • In this and th« following examples the magnitudes only of velocity and acceleration are required. 108 DIFFERENTIAL CALCULUS 9. Given a = 2 1 + 8 1> + 4 f* f t. ; find velocity and acceleration (a) at origin ; (b) at end of 6 seconds. Ana. (a) o = 2 ft. per sec., a = 6 f t. per (sec.)^ ; (b) c = 832 ft per sec., a = 126 ft per (sec.)«. 10. Given s = - + bf^, where a and b are constants ; find velocity and accelera- tion at any instant Ana. v = h2M, a = \-2b. 11. At the end of t seconds a body has a velocity of 3 ^ + 2 1 ft per sec. ; find its acceleration (a) in general ; (b) at the end of 4 seconds. ^718. (a) a = 6 1 + 2 ft per (sec.)^ ; (b) a = 26 ft. per (sec.)>. 12. The vertical component of velocity of a point at the end of t seconds is «y = 8«? - 2f + 6 ft per sec. Find the vertical component of acceleration (a) at any instant ; (b) at the end of 2 seconds. Ana. (a) a, = 6t — 2 ; (b) 10 ft. per (sec.)^. 13. If a point moves in a fixed path so that a = Vt, show that the acceleration is negative and proportional to the cube of the velocity. 14. If the distance in feet described by a point in t seconds is given by the formula « = 101og , 4 + t find velocity and acceleration (a) at the end of 1 second ; (b) at the end of 16 seconds. Ana. (a) c = — 2 ft per sec, o = J ft per (sec.)^; (b) r = — i ft per sec, o = ,^ f t. per (sec.)*. 16. If the space described is given by show that the acceleration is always equal to the space passed over. 16. Given s = a cos — : find acceleration. Ana. a = 2 4 17. If a point referred to rectangular coordinates moves so that X = a cos £ + &, and y = a sin f + c, show that its velocity has a constant magnitude. 18. If the path of a moving point is the sine curve (x = a^, ( y = 6 sin o^, show (a) that the x component of the velocity is constant ; (b) that the acceleration of the point at any instant is proportional to its distance from the axis of X. 19. If a particle moves so that « = r», y = <«, (a) show that the path is the semicubical parabola y^ = x^; (b) find Vxy v^, V ; (c) find Ox, o^, a ; (d) when t = 2 sec, find v, a, position of point (coordinates), and direction of motion. CHAPTER Vin SUCCESSIVE DIFFERENTIATION 87. Definition of successive derivatives. We have seen that the derivative of a function of x is in general also a function of a*. This new function may also be differentiable, in which case the derivative of the jir%t derivative is called the second derivative of the original function. Similarly the derivative of the second derivative is called the third derivative; and so on to the nth derivative. Thus, if dx dx\axj dx\jdx\dxj^ 88. Notation. The symbols for the successive derivatives are usually abbreviated as follows : . dx\dxj do(^ dx\_dx\dx)^ dx\dx'J da? d_ /d'-'yN ^ ^ dxydx'-y dxr' If 1/ =f(T), the successive derivatives are also denoted by /'(^). /"(^), f"'{^), n^)^ ■■; r'H^)i f/^^)' £-^<^>' y^'^^ £^-^(^)' •••' ^•^<^>' 109 110 DIFFERENTIAL CALCULUS 89. The lyth derivative. For certain functions a general expres- sion involving n may be found for the nth derivative. The usual plan is to find a number of the first successive derivatives, as many as may be necessary to discover their law of formation, and then by induction write down the nth derivative. Ex. 1. Given y = e^i find — . Solution. — = oc"*, dx — i = a'^e^. Ana. Ex. 2. Given y = logx ; find dx» d*y dx* Solutum. — = -i dz X dx^ x^ ' d»y_l-2 dx' x« d^ _ 1 . 2 » 3 dx*" X* ' • • • • d"i/ In — 1 ^ = (-1)— 1!=. Am. dx" x» Ex. 3. Given v = sin x ; find — - . dx» Solution. — = cos X = sin (X + - ) dx \ 2/ = — sinfx + — ) = C08( X + - ) = Bin(x H — -\ dz \ 2/ \ 2/ \ 2 / dxa »y d . / , 2t\ / , 2t\ . / , 3t\ = -sini X H I = COS! X H 1 = smi x H li c«dx \2/ \2/ \2/ -^ = ami X H ). Ana. dx« \ 2 / SUCCESSIVE DIFFERENTIATION 111 90. Leibnitz's formttla for the nth derivative of a product. This formula expresses the nth derivative of the product of two variables in terms of the variables themselves and their successive derivatives. If u and V are functions of Xj we have, from Y, d , . du ' dv dx dx , dx Differentiating again with respect to Xy - cP , . cPiA du dv , du dv dFv — iuv) = V H 1 h u — d^? da? dx dx dx dx do? ePw , c%du dv ^ ePv da? ^ dxdx^ dj? Similarly d* , . d?u . dSi dv . rt d?u dv , g. du d?v du d'v , d?v — (wt;) = — V H 1- 2 h 2 1 h u — do? da? da? dx da? dx dx do? dx da? da? d?u . „ d?u dv „ du d?v d?v = — V '\' 6 h o h u — • da? da? dx dx da? da? However far this process may be continued, it will be seen that the numerical coefficients follow the same law as those of the Binomial Theorem, and the indices of the derivatives correspond to the exponents of the Binomial Theorem.* Reasoning then by mathematical induction from the wth to tlie (m + l)th derivative of the product, we can prove Leibnitz^ % Formula ^ ' dx^^^^^ ■" dx^ ^^'^ dx^-^ dx "^ [2 €te~-« cte« "•■ ' " du d^-^v ', d'^i^ ^ dxdx^'^^ dxn Ex- 1. Giyen y = e* log x ; find — ^ by Leibnitz^s Formula. SolutiOTi, Let tt = e*, and V = logx; then dv 1 dx " X dx^ cPc 1 dx2" X«* dhi dH 2 — = c*, — = — . dx' dx* X? * Tb make thii oorrespondenoe complete, u and v are oonaidered as —- and •-— < 112 DIFFERENTIAL CALCULUS Substitating in (17), we get 8 = ^(logx + --- + -) d"v Ex. 2. Given y = ««e«*; find — ^ by Leibnitz's Formula. Solution. Let tt = «», and c = e«« ; then d^ dx d^ d'o dx« d"u dx» d*^ dx"^ Substituting in (17), we get c^y — T ^«« «e4- ^nnn- -Wfox 4- n^n _ dx" ' = a»-«c«*[x«a« + 2nax + n(n-l)]. 91. Sttccessive differentiation of implicit functions. To illustrate the process we shall find -— ^ from the equation of the hyperbola 6 V - ay = a*6». Differentiating with respect to a: as in § 75, p. 84, 2 6»a:-2aVl^ = 0, or, M) ^ = ^ Differentiating again, remembering that y is a function of x^ ^__ dx SUCCESSIVE DIFFERENTIATION 113 dy Substituting for -^ its value from (-4.), epy ^ \a'y) ^ V{h'2?-^a^f) da? a^y^ aV But from the given equation, 6 V — ay = (j?h\ da? a^y* EXAMPLES Differentiate the following. 1. y = 4a!*-6x2 + 4x + 7. ^ = 12(2x-l). ox* 2. f{x) = ^. /iT(x) = 1 - X " ' ' (1 - X)* 3. /(y) = j^. /^(y)=l6. . 4. y = x« log X. d^y 6 da:* X K _ c ^ — n(n-f l)c 6. y = (x-3)^* + 4xe» + x. -4 = 4e'r(a; - 2)c* + x + 2] a. - --V (T^ 1 , - --, y 2^ ' dx^ 2a^ ' a» 8. /(x) = 0x9 + 6x + c. 9. /(x) = log(x + l). 10. /(x) = log(e* + e-*). 11. r = sin o^. 12. r = tan *. — - = 6 sec* — 4 sec* 0. d0« 13. r = log sin 0. — = 2 cot cosec' 0. (i0' 14. f(t) = e-'co8«. /iv(Q = -. 4e-'cos« = - 4/(t). dx2 2a^ ; - r'w = 0. /*Ma^) = 6 (X + 1)* r'(x) = 8 (c - e- x)8 de* a* sin od = a*r. 15. f{e) = V8ec2^. /" (^) = 3 [f(0)y - f(0). 114 DIFFERENTIAL CALCULUS 16. p = (g« + a2)arcUn^ ' - a dq* (a=» + q^* 17. y = a*. 3^ = (log a)» a*. 18. y = log(l+«). dx« d»v |n-l rii? = (-i)«-i ' , d*i/ / nT \ 19. v=rco8ax. -~ = a»cosiaxH )• dx" \ 2 / 20. y = x»-Uogx. -i^ = i=. OX'* X [nm a poeitive integer.] 21. y = l=^. ^ = 2(-l)«— i? ^ 1 + x 'dx» ^ ' (l + a)"+* 2 Hint. Reduce fraction to form - 1 + z before differentiating. 22. If y = e=^8inx, prove that ~|-2-^ + 2y = 0. dx^ dx dhf dy 23. Uy=z acoB (log x) + 6 sin (log x), prove that x^— ~H-x~ + y = (X dx^ dx Use Leibnitz^s Formula in the next four examples. 24. y = x«a*. t^ = <»* (log a)""' [(« log a + n)J» - n]. dx" d*y 26. y = e*ix. --^ = c*(x + n). dx*» •26. /(x) = c«sinx. /(")(x) = (V2)'«e«8in(x + — )• 27. f{e) = cos o^ COB be. /C»>(^) = <?±^ cos [(a + 6) ^ + — 1 , (a - 6)" r . , . _ , nr"! + ^ COS [^(a - 6) ^ + _ J . 28. Show that the formulas for acceleration, (14), (15), p. 105, may be written _d2« _ _d*x _ _d^ """d£2 29. y« = 4ax. 30. Wx2 4. a2y3 = a2ft2. 31. xa + y« = A 32. y^ + y=:x*. dx« (l + 2y)* df'i' " d«2* d2y 4a* dx* y» d»y _ &* . d«y _ S¥x dxa~ a2y8' dx«~ aV d2y t^ dx* y« d»y 24 X SUCCESSIVE DIFFERENTIATION 115 33. ax«H-2Axy + 6y« = l. 38. €« + tt = e" + «. 41. y« + «'-8axy = 0. (Py_ h^- ab dxa "" (Ax + by)* 34. ya-2xy = a«. _=__; _ = -__. cP^ tana e - ten* d02 tan* $ d^e 2(6 + 8^ + 8^) 36. tf = ten(0 + (^). 5^ = --^^ ^5 cPc 4(tt + t)) 37. log(u + r) = u-t,. _ = ___. 39. a = l + «e.. _ = __^-. <J2« (2-a)c + 2« 40. c" + rt-c = 0. 3^5 = * dp (C + 0» d^y _ 2 a»zy ^■" (y«-ax)»' d»y a(m«-l) 42. y.-2»«y + ««-a = 0. _ = ^_--^,. d?y — y 43. y = 8in(x + y). Tl = 7; ; — i — ^15* 9 \ "f dx» [1 — cos (X + y)]» d^_ y[(x-l)« + (y -!)»] . 44. e»+. = xy. ^= ^^^-^ji 46. ax« + 2Axy + 6y« + 2fifX4-2/y + c = 0. CHAPTER IX MAXIMA AND MINIMA* 92. Increasing and decreasing functions. A function is said to be increaBing when it increases as the variable increases and decreases as the variable decreases. A function is said to be decreasing when it decreases as the variable increases and increases as the variable decreases. The graph of a function indicates plainly whether it is increas- ing or decreasing. For instance, consider the function a* whose graph (Fig. a) is the locus of the equation y = a*. a > 1 As we move along the curve from left to right the curve is m- ing^ i.e. as x increases the function (= y) always increases. There- fore a* is an increasing function for all values of x. Y Ay X Fio. a Flo. 5 On the other hand, consider the function (a — x)* whose graph (Fig. h) is the locus of the equation y = (« - ^f- * The proofs given in this chapter depend chiefly on geometric Intuition. The sul^ect of Maxima and Minima will he treated analytically in $ 119, p. 169. 116 MAXIMA AND MINIMA 117 Now as we move along the curve from left to right the curve is falling^ i.e. as x increases the function (= y) always decreases. Hence (a — a;)' is a decreasing function for all values of x. That a function may be sometimes increasing and sometimes decreasing is shown by the graph (Fig. c) of y=2i»-9a? + 12a;-3. As we move along the curve from left to right the curve lises until we reach the point A^ then it falls from ^ to ^, and to the right of B it is always rising. Hence (a) from re = — oo to a; = 1 the function is increasing; (b) from x=zl to a: = 2 the function is de- creasing; (c) from x=2 to a? = -foo th^ function is increasing. The student should study the curve care- fully in order to note the behavior of the fio. « function when a? = 1 and a; = 2. Evidently A and B are turning points. At A the function ceases to increase and commences to decrease ; at B, the reverse is true. At A and B the tangent (or curve) is evidently parallel to the axis of X, and therefore the slope is zero. 93. Tests for determining when a function is increasing and when decreasing. It is evident from Fig. c that at a point, as C, where a function ^. ^.^ is increasing^ the tangent in general makes an acute angle with the axis of X; hence slope = tan t = -~ =/' (x) = a positive number. dx Similarly at a point, as 2>, where a function is decreasing^ the tan- gent in general makes an obtuse angle with the axis of X\ therefore slope = tan t = 3^ =/' {x) = a negative number,* dx * ConTeraely, for any glren ralue of x, \ffXx)~+f then f{x) %9 incretMtng ; iff{z)='- , tken/(x) U decreasing . Wben/*(z)=0, we cannot decide without further InTestigation whether /(x) is increasing or decreasing. 118 DIFFERENTIAL CALCULUS In order then that the function shall change from an increasmg to a decreasing function, or vice versa, it is a necessary and suffi- cient condition that the first derivatiye shall change sign. But this can only happen for a continuous derivative by passing through the value zero. Thus in Fig. c, p. 117, as we pass along the curve the derivative (= slope) changes sign at A and B where it has the value zero. In general then we have at turning points (18) ^=/r(x) = 0. The derivative is continuous in nearly all our important appli- cations, but it is interesting to note the case when the derivative (= slope) changes sign by passing through oo.* This would evi- dently happen at the points B^ E, G in Fig. d, p. 119, where the tangents (and curve) are perpendicular to the axis of X At such exceptional turning points or, what amounts to the same thing, 1 /(^) = 0. 94. Maximum and minimum values of a function, f A maximum value of a function is one that is greater than any values imme- diately preceding or following. A minimum value of a function is one that is less than any values immediately preceding or following. For example, in Fig. c, p. 117, it is clear that the function has a maximum value MA (= y = 2)^when a? = 1, and a minimum value NB{=i/= 1) when x=2. The student should observe that a maximum value is not neces- sarily the greatest possible value of a function nor a minimum value the least. For, in Fig. c it is seen that the function (= y) has values to the right of B that are greater than the maximum MA^ and values to the left of A that are less than the minimum NB. * By this is meant that its reciprocal passes tlirouKli the value zero. t The student should not forget that in general the definitions and proofs giren in this book apply only at points where the function is continuous. MAXIMA AND MINIMA 119 A function may have several maximum and minimum values. Suppose that the following figure represents the graph of a function f{x). At B^ Z>, G, /, K the function is a maximum, and at C, E^ jfiT, J a minimum. That some particular minimum value of a function may be greater than some particular maximum value is shown in FlO. d the figure, the minimum values at C and H being greater than the maximum value at K. At the ordinary turning points C, 2>, H^ /, Ji K the tangent (or curve) is parallel to OX; therefore «Zoj9e = ^=/'(2r) = 0. At the exceptional turning points B^ E^ G the tangent (or curve) is perpendicular to OX, giving %lope = ^ = dx f{x) = oo. One of these two conditions is then necessary in order that the function shall have a maximum or a minimum value. But such a condition is not sufficient ; for, at F the slope is zero and at A it is infinite, and yet the function has neither a maximum nor a minimum value at either point. It is necessary for us to know, in addition, how the function behaves in the neighborhood of each point. Thus at the points of maximum value^ B, Z>, Gj /, K, the function changes from an increasing to a decreasing function^ and at the points of minimum value^ C, E^ if, Ji the function changes 120 DIFFERENTIAL CALCULUS frtym a decreasing to an increasing function. It therefore follows from § 93 that at maTimum points slope = 3^ =f'(x) must change from + to — » dx and at minimum points slope =-^ =f\x) must change from — to 4 when we move along the curve from left to right. At such points as A and i^ where the slope is zero or infinite, but which are neither maximum nor minimum points^ slope = -~- =/'(2:) does not change sign, dx We may then state the conditions in general for maximum and minimum values oif{x) for certain values of the variable as follows : (19) f(x) is a maximum if /'(x) = O, and f(Qc) chansres from + to — . (20) f(x) is a minimum if /'(x) = O, and f^(x) chansres from — to +• The values of the variable at the turning points of a function are called critical values; thus x=\ and x = 2 are the critical values of the variable for the function whose graph is shown in Fig. (7, p. 117. The critical values at turning points where the tangent is parallel to OX are evidently found by placing the first derivative equal to zero and solving for real values of x^ just as under §77, p. 86.* To determine the sign of the first derivative at points near a particular turning point, substitute in it, first a value of the variable just a little less than the corresponding critical value, and then one a little greater.^ If the first gives + (as at X, Fig. rf, p. 119) and the second — (as at M)^ then the function (= y) has a maximum value in that interval (as at /). * Similarly if we wish to examine a function at exceptional turning points where the tangent is perpendicular to OX, we set the reciprocal of the first deriyatire equal to aero and solve to find critical values. t In this connection the term " little less," or " trifle less,** means any value between tlio next smaller root (critical value) and the one under consideration ; and the term " little greater,*' or " trifie greater," means any value between the root under consideration and the next larger onck MAXIMA AND MINIMA 121 If the first gives — (as at P) and the second + (as at iV), then the function (= y) has a minimum value in that interval (as at C). If the sign is the same in both cases (as at Q and JS), then the function (= y) has neither a maximum nor a minimum value in that interval (as at F).* We shall now summarize our results into a compact working rule. 95. First method for examining a function for maximum and minimum values. Working rule. First step. Find the first derivative of the function. Second step. Set the first derivative equal to zero f and solve the resulting equation for real roots in order to find the critical values of the variable. Third step. Write the derivative in factor form; if it is algebraic^ write it in linear factor form. Fourth step. Considering one critical value at a iimcy test the first derivative^ first for a value a trifle less and then for a value a trifle greater than the critical value. If the sign of the derivative is first -f and then — , the function has a maximum value for thai partic- ular critical value of the variable; but if the reverse is true^ then it has a minimum value. If the sign does not change^ the function has neither. Ex. 1. Examine the function (x — 1)^(2 + 1)' for maximnm and minimum values. SoLxdvoTL f(z) = (X - 1)« (X + 1)». Firststep. /'(x)=2(x-lXa; + l)H3(x~l)2(x + l)2 J = (X - l)(x + l)«(6x - 1). ..i^X Second step. (x-l)(x+ 1)«(6» -1)=0, ^ X = 1, — 1} i, whicli are critical yalues. Third step. /'(x) = 5(x - 1) (x + l)«(x - i). Fourth step. Examine first for critical value x = 1 (C in figure). Whenx<l,r(x) = 5(-) (+)«(+) = -. When X > 1, r{x) = 5(+)(+)«(+) = +. Therefore, when x = 1 the function has a minimum value /(I) = (= ordinate of C). * A similar disctusion will evidently hold for the exceptional turning points £, E, and A respectlrely. t When the first derlTatire becomes infinite for a certain ralue of the independent variable, then the function should be examined for such a critical value of the variable, for it may give maximum or minimum values, as at B, E, or A (Fig. d, p. 119). See footnote on preceding page. 122 DIFFERENTIAL CALCULUS Examine now for critical value x = | (B in figure). When x<i, r(x) = 6(-) (+)2(-) = +• Whenx>J,/'(x) = 5(-)(+)2(+)=-. Therefore, when x = J the function has a maximum value /(}) = 1.11 + (= ordi- nate of B). Examine lastly for critical value x = — 1 (^ in figure). Whenx<-l,/'(x) = 6(-)(-)5»(-) = +. When X > - 1, r(x) = 5(-)(+)2(-) = +. Therefore, when x = — 1 the function has neither a maximum nor a minimum value. Ex. 2. Examine sin x (1 + cos x) for maximum and minimum values. Solvtion. f{x) = sin x (1 + cos x). First step. f'(x) = -sin2x-|-(l + cosx)co8X=2coa^x + coBX- 1. Second atep. 2 cos^ x + cosx — 1 = 0. Solving the quadratic, cos x = J or — 1 ; hence the critical values are x = i: - or t. 3 Third step. f\x) = 2 (cos x - i) (cos x + 1). Fourth, step. Examine first for critical value x = — 3 When x<|, /'(x) = 2 (+)(+)=+. When x>^, /'(x) = 2 (-)(+) = -. Therefore, when x = ^ the function has a maximum value /(-) = - VS. 3 \3/ 4 Examine now for critical value x = • 3 Whenx<-^, /'(x) = 2 (-)(+) = -. When x> - ^, f\x) = 2(+)(+) = +. Therefore, when x = - - the function has a minimum value /(--) = — Vs. 3 V 3/ 4 Examine now for critical value x = ir. When x<x, /'(x) = 2 (-)(+) =-. Whenx>x, /'(x) = 2 (-)(+)= -. Therefore, when x = «- the function has neither a maximum nor a minimum value. MAXIMA AND MINIMA 123 Since the cosine is a periodic function, the critical yalues are really X = 2 nir ± - and nr, where n is any integer. Therefore the function has an infinite number of maxima all equal to ] V3, and an infinite number of minima all equal to — } V3. £x. 3. Examine the function a - 6 (x — c)' for maxima and minima. Solution. f(x) = a — 6(x — c)K 3(x-c)* Since x = c is a critical value for which /'(x) = 00, but for which /(x) is not infinite, let us test the function for maximum and minimum values when x = c. When x<c, /'(x) = +. When X > c, /'(x) = - . Hence, when x = c = OM the function has a maximum value /(c) = a = MP, 96. Second method for ezamining a function for maximum and minimum values. From (19), p. 120, it is dear that in the vicinity of a maximum value of /(a:), in passing along the graph from left f\x) changes from -{- to to ^,* Hence /'(x) is a decreasing function, and by § 93 we know that its derivative, i.e. the second derivative of the function itself [=/"(a:)], is negative or zero. Similarly we have, from (20), p. 120, that in the vicinity of a minimum value of f{x) /'(a:) changes from — to to +. Hence f\x) is an increasing function, and by § 93 it follows that f'\x) is positive or zero. The student should observe that /"(a:) is positive not only at minimum points (as at A) but also at points such as P. For, as a point ^ passes through P in moving from left to right, slope = tan t = -~ =f(x) is an increasing function. dx At such a point the curve is said to be concave upwards. *f{x) is aasumed to be contintiouB, and/^(x) to exist. 124 DIFFERENTIAL CALCULUS Similarly /"(x) is negative not only at maximnm points (as at B) but also at points such as Q. For, as a point passes through Q, glope = tan t = -~ =/' (x) in a deerea9inff function, dx At such a point the curve is said to be concave downwards,* We may then state the sufficient conditions for maximum and minimum values of f{x) for certain values of the variable as follows : (21) /(x) is a nuudmum if ff(x) = O and ff^(x) = a nefirative number. (22) f(x) is a minimum if f^(x) = O and /^^(x) = a positive number. Following is the corresponding working rule. First step. Find the first derivative of the function. Second step. Set the first derivative equal to zero and solve the resulting equation for real roots in order to find the critical values of the variable. Third step. Find the second derivative. Fourth step. Substitute each critical value for the variable in the second derivative. If the result is negative^ then the function is a maximum for that critical value ; if the result is positive^ the function is a minimum, jf Ex. 1. Examine x* — 3 x^ — z + 6 for maxima and minima. SolutUm. /(x) = x«-3x«-9x + 6. First step. f(x) = 3x« - 6x - 9. Second step. 8x* - 6x - 9 = 0; * hence the critical values are x = — 1 and 3. Third step. /"(x) = 6x-6. Fourth step. /''(-I) = -12. .-./(-I) = 10 = (ordinate of A) = maximum value. /" (3) = + 12. .•./(3) = - 22 (ordinate of B) = minimum value. * At a point where the cuxre is concave upwards tre sometimes say that the carve has & positive bending^ and where it is concave dotontvarda a negative bending. t When/"(x)sO, or does not exist, the above process fails, although there may even then be a maximam or a minimum ; in that case the first method given in the last section still holds, being fundamental. Usually this second method does apply, and when the process of flniUnq the second derivative is not too long or tedious, it is generally the shortest method. MAXIMA AND MINIMA 126 Ex. 2. Examine sin' x cosx for maximum and wtinimnm valaes. SolutioTL f{z) = sin^ z cos x. Fint step. f(z) = 2 sin z co8» x — sin' x. Second step, 2 sin x cos^ x — sin" x = ; "^ hence the critical values are x = nr ^ and X = nr i: arc tan VS = nx ± o. Third step, f* (x) = cos x (2 cos^ x - 7 sin« x). Fouxik step, /" (0) = +. .'./(O) = = minimum value at 0. /"■ (x) = — . .'. /(ir) = = maximum value at C. /" (o) = — . .-. /(o) = maximum value at A, /" (x — a) = + . .*. /(x — o) = minimum value at B, etc. The work of finding maximum and minimum values may frequently be simpli- fied by the aid of the following principles which follow at once from our diroussion of the subject. (1) The maximum and minimum values of a continuous function must occur aUernatdy, (2) When c is a positive constant, cf(z) is a maximum or a minimum for such values of X, and such only, as make f{z) a maximum or a minimum. Hence, in determining the critical values of x and testing for maxima and minima, any constant factor may be omitted. When c is negative, cf{z) is a maximum v)henf(z) is a minimum, and conversely, (3) Ifc is a constant, -, . , , ^, . ^ ' f(x) and c +/(x) have maximum and minimum values for the same values ofz* Hence a constant term may be omitted when finding critical values of x and testing. Examine the following functions for maximum and minimum values. 1. 8x«-9x«-27x + 30. Arvi, x = - 1, gives max. =45; X = 3, gives min. = — 61. 2. 2x«-21xa + 36x-20. AnA, x = 1, gives max. = - 8 ; X = 6, gives min. = — 128. x" 3. 2x* + 3x + l. • An», x = 1, gives max. ={; 3 X = 8, gives min. = 1. 4. 2x»-16x« + 86x + 10. An», x = 2, gives max. = 38 ; X = 3, gives min. = 37. 5. x* — 9x* + 15x — 3. An», x = 1, gives max. = 4 ; X = 5, gives min. = — 28. 8. x* — 3x^ + 6x + 10. Anjs, No max. or min. 7. X* ~ 5x* + fix* + 1. An», x = 1, gives max. = 2 ; X = 3, gives min. = — 26 ; X = 0, gives neither. 126 DIFFERENTIAL CALCULUS 8. 3z»~ 126x8 + 2160X. 9. (x-3)2(x-2). 10. (x-l)8(x-2)2. 11. (z - 4)6 (X + 2)*. 12. (x-2)6(2x+l)*. 13. (x + l)'(x-6)2. 14. (2x-a)*(x-a)l 15. «(x-l)«(x + l)«. 16. X (a + x)« (a - x)«. 17. 6 + c(x -a)'. 18. a-b(x-c)K 19. ^^-7x + 6 20. 21. 22. 23. X - 10 • (a- a - -X)8 2x 1- z + x« 1 + x-x* X2- ■3x + 2 x2 + 3x + 2 {X- -a)(6-x) X2 ^2. 6a X a — X ^na. X = — 4 and 3, give max. ; X = — 8 and 4, give mln. Am, X = }, gives max. = ^ ; X = 3, gives min. = 0. Aivi, X = j, gives max. = .03456 ; X = 2, gives min. = ; X = 1, gives neither. Am. X = — 2, gives max. ; X = f , gives min. ; X = 4, gives neither. Ai\», X = — ^, gives max. ; X = II, gives min. ; X = 2, gives neither. Am, X = If gives max. ; X = — 1 and 6, give min. 2a Am, X = — f gives max. ; X = a, gives min. ; X = -» gives neither. 2 Ana. X = ^, gives max. ; X = 1 and — I, give min. An8. X = — a and - , give max. ; o X = — » gives mm. 2 Ana. X = a, gives min. = 6. Am, No max. or min. Am. X = 4, gives max. ; X = 16, gives min. Am. X = - f gives min. 4 Ana. X = }, gives min. Ana. X = V2, gives min. = 12 V2 — 17 ; x=— V2, givesmax. = — 12v^-17. . 2ab (a-6)a Am. X = ri gives max. = - - - a + h 4a6 Ana. x = a'^ « = a -- 6 a + 6 t gives min. ; t gives max. 26. X logx 26. X MAXIMA AND MINIMA 127 Ans. X = e, gives min. Ana. MiD. when x lies between ^f and |}. 27. ae** + be-**. Ana. Min. = 2 VoS. 28. X'. Ana. x = -y gives min. e 29. x^. Ana. x = e, gives max. 30. COB X + sin X. Ana. x = - » gives max. = V2 ; 4 X = — - , gives min. = — V2. 4 31. sin 2 X — X. ^n«. x = - 1 gives max. ; X = ~ — I gives min. 6 32. X + tan x. ^n«. No max. or min. T 3 /- 33. sin^x cos x. ^tu. x = - , gives max. = — v3 : 3 ^ 16 r . . 3 /r X = — -- , gives mm. = VS. 3 * 16 34. xcosx. ^715. X = cot X, gives max. 36. sin X + cos 2 x. Ana. x = arc sin |, gives max. ; X = - , gives min. 36. 2 tan X — tan^x. Ana. x = - ) gives max. 4 -_ sinx A ^ • 37. . j^ns. x = —i gives max. 1 + tanx 4 X 38. Ana. x = cos x, gives max. 1 + xtanx 39. The range OA of a projectile in a vacuum is given by the formula _, ri2sin20 Ji = ; 9 where Vi = initial velocity, g = acceleration due to grav- ity, ^ = angle of projection with the horizontal. Find the angle of projection which gives the greatest range o for a given initial velocity. Ana. = 46®. 40. The total time of flight of the projectile in the last problem is given by the formula _ 2tJi8m0 J^ = . 9 At what angle should it be projected iu order to make the time of flight a maximum ? Ana. 4> = 90®. 128 DIFFERENTIAL CALCULUS 41. The time it takes a ball to roll down an inclined plane AB is giyen by the a formula ^^^ r if08m2^ or sin 20 Neglecting friction, etc., what must be the yalue of to make the r^ quickest descent ? Aia. ^ = 46°. 42. When the resistance of air is taken into account, the inclination of a pen- dulum to the vertical may be given by the formula $ = ac-** cos (tU + e). Show that the greatest elongations occur at equal intervals - of time. n 43. It is required to measure a certain unknown magnitude x with precision. Suppose that n equally careful observations of the magnitude are made, giving the results ai, Os} da* • • • » flu* The errors of these observations are evidently X — fli, X — Osi X — Os, • • • , X — Oa, some of which are positive and some negative. It has been agreed that the most probable value of x is such that it renders the sum of the squares of the errors, namely (X - ai)« + (X - a,)«+ (X - aa)a + . . . + (x - a,)«, a minimum. Show that this gives the arithmetical mean of the observations as the most probable value of x. 44. The bending moment at B of a beam of length I, uniformly loaded, is given by the formula 3f=i«rfx-lwx?, where w = load per unit length. Show that the maximum bending moment is at the center of the beam. 45. If the total waste per mile in an electric conductor is r where c = current in amperes, r = resistance in ohms per mile, and f = a constant depending on the interest on the investment and the depreciation of the plant, what is the relation between c, r, and t when the waste is a minimum ? Ans. cr = t, 46. A submarine telegraph cable consists of a core of copper wires with a cover- ing made of nonconducting material. If x denote the ratio of the radius of the core to the thickness of the covering, it is known that the speed of signaling varies as xMog-. X Show that the greatest speed is attained when x = — • MAXIMA AND MINIMA 129 47. Afisaming that the power given out by a voltaic cell Is given by the formula where E = constant electro-motive force, r = constant internal resistance, B = exter- nal resistance, prove that P is a maximum when r = R, 48. When a battery of mn cells is joined up so that m rows of n cells, connected in series, are joined in parallel, the current is given by the formula _ fiinE ^ {j =r — — , mR-^-wr where E = electro-motive force of each cell, r = internal, and S = external resist- ance of each cell. Show that the current is a maximum when Rm = m, that is, the total internal resistance equals the total external resistance. 49. The force exerted by a circular electric current of radius a on a small magnet whose axis coincides with the axis of the circle varies as X where x = distance of magnet from plane of circle. Prove that the force is a maxi- mum when x = -• 2 60. We have two sources of heat at A and B with intensities a and h respectively. The total intensity of heat at a distance of x from A is given by the formula 1 = -^+ * Show that the temperature at P will be the lowest when -4- F S d-x _Vb th^t is, the distances BP and AP have the same ratio as the cube roots of the corresponding heat intensities. The distance of P from A is aki x= a* + 6* 97. General directions for solving problems involving maxima and minima. In our work thus far the function has been given whose maximum and minimum values were required. This is by no means always the case ; in fact, we are generally obliged to construct the function ourselves from the conditions given in the problem, and then test it as we have been doing for maximum and minimum values. No rule applicable in all cases can be given for constructing the function, but in a large number of problems we may be guided by the following 130 DIFFERENTIAL CALCULUS General Directions. (a) Express the function whose maodmum or minimum i% involved in the problem, (6) If the resulting expression contains more than on^ variable^ the conditions of the problem will furnish enough relations between the variables so that all may be expressed in terms of a single one, {c) To the resulting functioii of a single variable apply one of our two rules for finding! maximum and minimum values, (d) In practical problems it is usually easy to tell which critical value will give a maximum and which a minimum value^ so it is not always necessary to apply the fourth step of our rules. m PROBLEMS 1. It is desired to make an open-top box of greatest possible volume from a square piece of tin wbpse side is a by cutting equal squares out of the comers and then folding up the tin to form the sides. What should be the length of a side of the squares cut out ? SolvJtion. Let x = side of small square = depth of box ; then a — 2 X = side of square forming bottom of I) box, and volume is F = (a-2x)«x; ^ which is the function to be made a maximum by varying x. Applying rule, dV First step. "T" = (« - 2x)2 - 4x(a - 2x) = a^ - 8ax + 12x«. ax Second step. Solving a^ — 8ax + 12x^ = gives critical values x = - and -• d It is evident from the figure that x = - must give a minimum, for then all the tin would be cut away, leaving no material out of which to make a box. By the usual a 2 a? test, X = - is found to give a maximum volume — • Hence the side of the square 6 27 to be cut out is one sixth of the side of the given square. 2. Assuming that the strength of a beam with rectangular cross section varies directly as the breadth and as the square of the depth, what are the dimensions of the strongest beam that can be sawed out of a round log whose diameter is d ? Sohdion, If x = breadth and y = depth, then the beam will have maximum strength when the function xy^ is a maximum. From the figure, y^ = d^ — x^ ; hence we should test the function /(x) = X ((P - x*). MAXIMA AND MINIMA 131 First step. f(x) = - 2x3 + (P - x« = d» - 8»a a Second step, (P— 3 x2= 0. .*. x = -—= = critical value which gives a maximum. V3 Therefore, M the beam is cat so that depth = Vf of the diameter of log, and breadth = Vi of the diameter of log, the beam will have maximum strength. 3. What is the width of the rectangle of maximum area that can be inscribed in a given segment OAA* of a parabola ? Hint. If OC-A, BC'^h-x and PP'=»2y; therefore the area of rectangle PDJyP' is 2{h-x)y. But since P lies on the parabola y* = 2px, the function to be tested is 2ih-x)y/2px. Ans, Width = 1^ 4. Find the altitude of the cone of maximum volume that can be inscribed in a sphere of radius r. Bint. Volume of cone- iira:»y. But x«= BCx CZ)« y (2r-y) ; there- fore the function to be tested is /(y)-^y'(2r-y). Ans, Altitude of cone = } r. 6. Find the altitude of the cylinder of maximum volume that can be inscribed in a given right cone. Bint. Let AC^ r and BC= h. Volume of cylinder « nafly. But from similar triangles ABC and DBO K ^ r{h-y) r:x::h:h~y, .*. x= — - — * A Hence the function to be tested is /(y)=y(*-y)*. Ans. Altitude = i A. 6. Divide a into two parts such that their product is a °>a^^^™- Ans. Each part = ?. 2 7. Divide 10 into two parts such that the sum of their squares is a minimum. Ans, Each part = 5. 8. Divide 10 into two such parts that the sum of the double of one and square of the other may be a minimum. -^'M. 9 and 1. 9. Find the number that exceeds its square by the greatest possible quantity. Ans. |. 10. What number added to its reciprocal gives the least possible sum ? Ans. 1. 11. Assuming that the stiffness of a beam of rectangular cross section varies directly as the breadth and the cube of the depth, what must be the breadth of the stiffest beam that can be cut from ft log 16 Uiches in diameter ? Ans. Breadth = 8 inches. 132 DIFFERENTIAL CALCULUS 12. A torpedo boat is anchored 9 miles from the nearest point of a beach, and it is desired to send a messenger in the shortest possible time to a military camp situated 15 miles from that pohit along the shore. If he can walk 5 miles an hour but row only 4 miles an hour, required the place he must land. An8» 3 miles from the camp. 13. For a certain specified sum a man takes the contract to build a rectangular water tank lined with lead. It has a square base and open top, and holds 108 cubic yards. What shall be its dimensions in order to require the least possible quantity of lead ? AuB. Altitude = 3 yds., side = 6 yds., that is, length of side = twice the altitude. 14. A gasholder is a cylindrical vessel closed at the top and open at the bottom, where it sinks into the water. What should be its proportions for a given volume to require the least material (this would also give least weight) ? Ana. Diameter = double the height. 16. What should be the dimensions and weight of a gasholder of 8,000,000 cubic feet capacity built in the most economical manner out of sheet iron ^ of an inch thick and weighing 2^ lbs. per sq. ft.? Ana. Height = 137 ft., diam. = 273 ft, weight = 200 tons. 16. What are the most economical proportions for a cylindrical steam boiler ? Ana. Diameter = length. 17. A paper-box manufacturer has in stock a quantity of strawboard 30 inches by 14 inches. Out of this material he wishes to make open-top boxes by cutting equal squares out of each comer and then folding up to form the sides. Find the side of the square that should be cut out in order to give the boxes maximum volume. Ana. 3 inches. 18. A roofer wishes to make an open gutter of maximum capacity whose bottom and sides are each 4 inches wide and whose sides have the same slope. What should be the width across the top ? Am. 8 inches. 19. Assuming that the energy expended in driving a steamboat through the water varies as the cube of her velocity, find her most economical rate per hour when steaming against a current running c miles per hour. Hint, Let v - most eoonomlcftl speed ; then av* « energy expended each hour, a being a constant depending npon the psrticalar oonditionB, and V - e » actual distance advanced per hour. at* Hence is the energy expended per mile of distance advanced, and it Is therefore the v-c function whose minimum is wanted. Ana. v = { c. 20. Prove that a conical tent of a given capacity will require the least amount of canvas when the height is V2 times the radius of the base. Show that when the canvas is laid out flat it will be a circle with a sector of about 66^ cut out A bell tent 10 ft. high should then have a base of diameter 14 ft. and would require 272 sq. ft. of canvas. MAXIMA AND MINIMA 133 21. Find the right triangle of maximum area that can be constructed on a line of length h as hypotenuse. . h , _., . ^ .». , ° •' *^ Ana. —rz = length of both legs. 22. Show that a square is the rectangle of maximum area that can be inscribed in a given circle. Also show that the square has the maximum perimeter. 23. What is the isosceles triangle of maximum area that can be inscribed in a given circle ? Ans. An equilateral triangle. 24. Find the altitude of the maximum rectangle that can be inscribed in a right triangle with base h and altitude fu .i « i... ^ ^ ** Ans. Altitude = -• 2 25. Find the dimensions of the rectangle of maximum area that can be inscribed in the ellipse li^ + aV = a^ft^. jin^, a Vi and 6 V2 ; area = 2a6. 26. Find the altitude of the right cylinder of maximum volume that can be inscribed in a sphere of radius r. , . ,^,^ j « 1. j 2 r ^ Ans. Altitude of cylinder = —-1 . Vs 27. Find the altitude of the right cylinder of maximum convex (curved) surface that can be inscribed hi a given sphere. ^^3. Altitude of cylinder = r V2. 28. Find the altitude of the right cylinder inscribed in a given sphere when its entire surface is a maximum. . ..^.^ , /» 2 \i Ana. Altitude = 12 — j r. 29. Find the altitude of the right cone inscribed in a sphere when its entire surface is a maximum. ^^ Altitude = (23 -VlT)^. 30. Find the altitude of the right cone of minimum volume circumscribed about a given sphere. Ana, Altitude = 4r, and volume = 2 x vol. of sphere. 31. A right cone of maximum volume is inscribed in a given right cone, the vertex of the inside cone being at the center of the base of the given cone. Show that the altitude of the inside cone is one third the altitude of the given cone. 32. Through a point (a, b) referred to rectangular axes a straight line is to be drawn, forming with the axes a triangle of least area. Show that its intercepts on the axes are 2 a and 2 b, 33. Through the point (a, b) a line is drawn such that the part intercepted between the axes is a minimum. Prove that its length is (a' + 5' )'. 34. Given a point on the axis of the parabola ^ = 2|»: at a distance a from the vertex ; find the abscissa of the point of the curve nearest to it. Ans. x = a — p. 35. What is the length of the shortest line that can be drawn tangent to the ellipse lAfl + oflffi = a^ and meeting the coordinate axes ? Ana. a + 6. 36. Find the altitude of the least isosceles triangle that can be circumscribed about an ellipse, the base being parallel to the major axis. Ana. Altitude = 3 times semi-minor axis. 134 DIFFERENTIAL CALCULUS 37. A Norman window consists of a rectangle surmounted by a semicircle. Given the perimeter; required the height and breadth of the window when the quantity of light admitted is a maximum. Ana. Radius of circle = height of rectangle. 38. A tapestry 7 feet in height is hung on a wall so that its lower edge is 9 feet above an observer's eye. At what distance from the wall should he stand in order to obtain the most favorable view ? Hint. The vertical angle Bubtended by the tapeetry in the eye of the obeerver must he at a maximum. An8. 12 feet 39. The regulations of the British Parcels Post require that the sum of the length and girth of a parcel shall not exceed 6 feet. Show that (a) The greatest sphere allowed is about 17} inches in diameter and has a volume of about 1^ cubic feet. (b) The greatest cube has an edge 14} inches in length and a volume of nearly 1| cubic feet. (c) The greatest rectangular box is 1 ft. by 1 ft. by 2 ft , containing 2 cubic feet (d) The greatest parcel of any shape is a cylinder 2 ft. long and 4 ft. cii'bum- ference, and contains over 2^ cubic feet 40. What are the most economical proportions of a tin can which shall have a given capacity, making allowance for waste ? r J ] Bint. There is no waste in cutting out tin for the side of the can, ^A^^A^.^^^,^^. but for top and bottom a hexagon of tin circumscribing the circular I 1^ I I P**®*" required is used up. ^o^^^*^'^^^*^^ Ana. Height = x diameter of base. r Note 1. Ex. 16 shows that if no allowance is made for waste, then height = diameter. Note 2. We know that the shape of a bee cell is hexagonal, giving a certain capacity for honey with the greatest possible economy of wax. 41. An open cylindrical trough is constructed by bending a given sheet of tin of breadth 2 a. Find the radius of the cylinder of which the trough forms a part when the capacity of the trough is a maximum. ^ _ , 2 a Ana. Rad. = — 42. A weight IT is to be raised by means of a lever with the force F at one end and the point of support at the other. If the weight is suspended from a point at a distance a from the point of support, and the weight of the beam is w pounds per linear foot, what should be the length of the lever in order that the force required to lift it shall be a minimum? SlTJr, , Ana. X = V feet ^ w 43. A rectangular stockade is to be built which must have a certain area. If a stone wall already constructed is available for one of the sides, find the dimensions which would make the cost of construction the least Ana. Side parallel to wall = twice the length of each end. M.VXT MA AND MINIMA 135 44. An electric arc light ta to be placed directly over the center of a ciicular plot o[ grass 100 feet [a diameter. Aasuming that the iDtenait; of light variea directly as tbe sine of the angle under which it strikes an illwninated surface and inversely as tbe square of its distance from the surface, how high should the liglit be hung in order that tbe best possible Ugbt shall fall on a walk along the circum- ference of the plot ? J 60 , . *^ Am. — T^feet v5 45. Tbe lower comer of a leaf, whose widUi is a, is folded over Eo as just to reach the inner edge of page, (a) Find the width of the part folded over when the length of the crease is a mioimiun. (b) Find width when the area folded over la a mEnimuin. Ar^. W?.i(b)?a. CHAPTER X POIKTS OF nVFIECTION 98. Definition. Potn^^c^f tn^6(?f ton separate arcs concave upwards from arcs concave downwards.* Thus, if a curve y =f{x) changes (as at B) from concave upwards (as at A) to concave downwards (as at C), or the reverse, then such a point as ^ is called a point of inflection. From the discussion of § 96 it follows at once that at A^ f\^) = +> and at C, f{x) = — . In order to change sign it must pass through the value zero;f hence we have (23) At points of in laection, ff(x)— O. Solving the equation resulting from (23) gives the abscissas of the points of inflection. To determine the direction of curving in the vicinity of a point of inflection, test f"{x) for values of a;, first a trifle less and then a trifle greater than the abscissa at that point. \if\x) changes sign, we have a point of inflection, and the signs obtained determine if the curve is concave upwards or concave downwards in the neighborhood of each point. The student should observe that near a point where the curve is concave upwards (as at A) the curve lies above the tangent, and at a point where the curve is concave downwards (as at C) the cxlrve lies below the tangent. At a point of inflection (as at B) the tangent evidently crosses the curve. • Points of inflection may also be defined as points where dhi d^ii (a) TT~ ^^'^ —^ changes sign, op oar* </a:« (b) — — • and — - changes sign. ay* dy* t It is assumed that fix) and/* (x) are continuous. The solution of Ex. 2, p. 137, shows how to discuss a case where f (x) and /"(x) are both infinite. Eyidently salient points (see p. 266; are excluded, since at such points /"(x) is discontinuous. 136 POINTS OF INFLECTION 137 Following is a rule for finding points of inflection of the curve whose equation is y =/(x), including also directions for examining the curve in the neighborhood of such a point. First step. Findf'\x). Second step. Setf^\x) = Oj and solve the equation for real roots. Third step. Write f\x) in factor form. . Fourth step. Test /"(x) for values of a:, first a trifle less and then a trifle greater than each root found in the second step. If f\Q^ changes sign^ we have a point of inflection, Whenf'\x)^'\-j the curve is concave upwards v^+^.* Whenf\x) = — , the curve is concave downwards y"'^^^^. EXAMPLES Examine the following curves for points of inflection and direction of bending. 1. y = 3a5*-4x« + l. Solution, f(x) = 3x*-4x« + l. First step. f'(x) = 36x2 - 24x. Second step. 36xa-24x = 0. .'. X = } and X = 0, critical values. Third step. f («) = 36 x (x ~ f). ¥<mrth step. When x < 0, /" (x) = + ; and when x > 0, /" (x) = — . .*. curve is concave upwards to the left and concave downwards to the right of x = 0{Am figure). When x< J,/"(x) = -; and when x > i,/"(x) =+. .*. curve is concave downwards to the left and concave upwards to the right of X = J (B in figure). The curve is evidently concave upwards everywhere to the left of -4, concave downwards between A (0, 1) and B (}, i^), and concave upwards everywhere to the right of B. 2. (y-2)« = (x-4). Solution, y = 2 + (X - 4)*. First step. ^ "^ S ^* " '*^" '' ^-.?(^_4)-t. dx2 9^ ' • This may be easily remembered if we say that a vessel shaped like the curve where it is ooncaye upwards holds (+) water, and where it is concave downwards spills (-) water. 138 DIFFERENTIAL CALCULUS Second step. When x = 4, both first and second deriyatives are infinite. Fourth step. Whenx<4, — ^ = + ; butwhenx>4, -i = -. dX* diXr We may therefore conclude that the tangent at (4, 2) is perpendicular to the axis of X, that to the left of (4, 2) the curve Lb concave upwards, and to the right of (4, 2) it is X concave downwards. Therefore (4, 2) must be considered a point of infiecUon. 3. V = 2'* -^y^- Concave upwards everywhere. 4. y = 5 — 2x — x^. Ans. Concave downwards everywhere. 5. y = x". Ans. Concave downwards to the left and concave upwards to the right of (0, 0). 6. y = x' — 8 x' — 9 X + 9. Ans. Concave downwards to the left and concave upwards to the right of (1, — 2). 7. y = a + (x — by. Ans. Concave downwards to the left and concave upvrards to the right of (6, a). jS 8. a*y = ax* + 2 a*. Ans. Concave downwards to the left and / 4 ox concave upwards to the right of f a, — j • 9. x« - 8 6x« + a«y = 0. Ans. Point of infiection is ( 6, — ^ J . 10. y = X*. Ans. Concave upwards everywhere. 11. y = x*-12x8 + 48x«-60. Ans. Concave upwards to the left of x = 2, concave downwards between x = 2 and x = 4, concave upwards to the right of x = 4. 12. y = -• Ans. Concave downwards between ( ± —-, — \ x2 + 4a« VV32/ concave upwards outside of these points. 13. y = x+36x2-2x»-x*. Ans. Points of infiection at x = 2 and - 3. x* / 9 a \ 14. y = — . Ans. Concave upwards to the left of I — 8 a, ) 1 x2 + 3a2 ^ \ 4 / concave downwards between T — 3 a, ) and (0, 0), concave upwards between (0, 0) and f 8 a, -j-)^ concave downwards to the right of f 8 a, — j • 16. (-) + (~) = 1. Ans. Points of inflection are x = ± -^. POINTS OF INFLECTION 139 16* oV = o^ - ^' -4iw. Points of inflection are x = ± - V27-3V38. 6 17. y = . Am. Points of inflection are x = 0, ±a VS. a« + x« 18. y = sin z. Aia. Points of inflection are x = nr, n being any integer. 19. y = tan x. Am. Points of inflection are x = rnr, n being any integer. 20. y = x€-'. Am, x = 2 gives a point of inflection. 21. Show that no conic section can have a point of inflection. 22. Show that the graphs of e* and log x have no points of inflection. 23. Show that the curve y (x^ + a') = x has three points of inflection lying on the straight line x — 4 €?y = 0. 24. Show that the abscissas of the points of inflection of the curve y^ = /(x) satisfy the equation [/'(x)]» = 2/(x).r(x). CHAPTER XI DIFFERENTIALS 99. Introduction. Thus far we have represented the deriyative of y =^f(x) by the notation We have taken special pains to impress on the student that the symbol dy dx was to be considered not as an ordinary fraction with rfy as numer- ator and dx as denominator, but as a single symbol denoting the limit of the quotient ^y Ax as Ax approaches the limit zero. Problems do occur, however, where it is very convenient to be able to give a meaning to dx and dy separately, and it is especially useful in applications of the Integral Calculus. How this may be done is explained in what follows. 100. Definitions. lif{x) is the derivative otf{x) for a particular value of a;, and Ax is an arbitrarily chosen increment of x^ then the dif- ferential off(x), denoted by the symbol df{x)^ is defined by the equation {A) df{x)=f{x)Ax. If now f(x) = a;, then f^{x) = 1, and (A) reduces to dx = Ax J showing that when x is the independent variable the differential of X {= dx) is identical with Ax. Hence, if y =f{x), (A) may in general be written in the form (B) dy=zff(x)dx.* * On account of the position which the deriTatiTeyCa;) here oocnples, it \b Bometime« called the d\ff'erenticU co^cient. The student should obserre the important fact that, since dx may be given any arbitrary yalue whatever, dx is independent qfx. Hence dy is a function qf two independent vcuiables x and dx. 140 DIFFERENTIALS 141 The differential of a function equals its derivative multiplied by the differential of the independent variable. y Let us illustrate what this means geomet- rically. 'Letf(z) be the derivative of y =/(a;) at P, Take dx = FQ, then dy=f{x)dx = tanr^FQ==^FQ=QT. M M'X Therefore rfy, or rff (a:), is the increment (= QT) of the ordinate of the tangent corresponding to dx.* This gives the following interpretation of the derivative as a fraction. If an arbitrarily chosen increment of the independent variable x for a point P (x^ y) on the curve y =f{x) be denoted by dx^ then in the derivative dx f (x) = tan T dy denotes the corresponding increment of the ordinate drawn to the tangent. 101. da: and dy considered as infinitesimals. In the Differential Calculus we are usually concerned with the derivative, i.e. with the ratio of the differentials dy and dx. In some applications it is also useful to consider dx as an infinitesimal (see § 30, p. 21). Then by (iS), p. 140, and (2), p. 27, dy is also an infinitesimal. Hence in such cases dx and dy are corresponding variables each of which approaches the limit zero.f 102. Derivative of the arc in rectangular coordinates* Let s be the lengthy of the arc AF measured from a fixed point A on the curve. Denoting the corresponding incre- ments by Ax^ Ay, As, we have from the figure (chord PQY=:(Axf^{Ay)\ * The student should note especially that thejdifferential {^dy) and the increment ("Ay) of the function corresponding to the same ralue dx{—Ax) are not in general equal. For, in the figure, dy =» QT but Ay = QP*. t In the Integral Calculus dx and dy are always regarded as infinitesimals. (Defined in §224. 142 DIFFERENTIAL CALCULUS Both multiplying and dividing the first member by (arc PQ)^ [=(A«)*], we get Dividing both members by (Ax)^ / chord Pg \YA«Y_j^ /AyV \aKPQ ) \i!ix) ^ \iix) ' From this we get, when Ax approaches the limit zero, ,1. , limit /chord P©\ ^ „ as8ummgthat^^^^(^— ^j = l. Hence (24) da doo --F^mf- Similarly, if we divide {A) by (Ay)" and pass to the limit, we get Also, from the above figure, cos 6 = Ax = ^._^^^,and chord P^ As chord P© chord P^ As chord Pg [Multiplying both numerator and denominator in each case by arc PQ (» A<).] As As approaches the limit zero, approaches the limit r, and the ratio of the arc P^ to the chord FQ approaches imity. Therefore (26) dx . dfi cos r = —- » sin r = -^* d8 da Using the notation of differentials, (24) and (25) may be written (98) -=[©■+']'•"• DIFFERENTIALS 143 An easy way to remember the relations (24) to (28) between the differentials dx^ dfy, ds is to note that they are correctly repre- sented by a right triangle whose hypotenuse is dsy sides dx and (2y, and angle at base r. Then , (i«=V(rfx)»4.(dy)«, and dividing by dx or rfy gives (24) or (25) respectively. Also, from the figure, / X V dz ^0 X cosr = -T-> sin ds ds the same relations given by (26). 103. Derivative of the arc in polar coordinates. In what follows we shall employ the same figure, notation, and reductions used on pp. 97, 98. From the right triangle PRQ (chord PQf = {PRf + (RQf = {p sin A^)* + (p + A/} — /} cos LOf = f? sin" A^ + (2 /) sin* ^ + Ap)". Multiplying and dividing the first member by (arc FQf [= (A»)^, and then dividing throughout by (A5)^ we get / chord Pe y/A^Y^ , / sin Ag V )( \KtzPQ ) \LB) ^\ t^e ) + \ />8in A^ «^°-2 A^ 2 ^A^ Passing to the limit as A0 diminishes towards zero (see § 80, p. 98), we get (S)'= (») ds d$ In the notation of differentials this becomes m -^<i-*oi''- 144 DIFFERENTIAL CALCULUS These relations between p and the differentials ds^ dp^ and d0 are correctly represented by a right triangle whose hypotenuse is d9 and sides dp and pd0. Then ds = ^{pddf 4. (dp)\ and dividing by d6 gives (89). Denoting by -^ the angle between dp and ds, we get at once de which is the same as {A)^ p. 98. 101 Formulas for finding the differentials of functions. Since . the differential of a function is its derivative multiplied by the differential of the independent variable, it follows at once that the formulas for finding differentials are the same as those for finding derivatives given in § 46, pp. 46-48, if we multiply each one by dx. This gives us I d!(c)=0. II d(x)=dx. III d(u + v — w)=du + dv — dw. IV d(cv)= cdv. V d(uv) = udv + »flJw. VI d (»,», ••■«,) = (v,v, •••»,) dv^ + («;,», . ■■vj)dv,+ + (»!"» • • '• »»-i) ^'v VII d(tf) = nif~^dv. Vila d(3f)= taf'^dx. VIII VIII a d c du — udv v" du Villi /c\ _ cdv DIFFERENTIALS 146 IX <i(lOgat^) = 10g„«— • V X d (a*") = a* log a dv, Xa d {e'') = e" dv. XI d{u'') = vu^'^du + log w . w" . cZw, XII c2 (sin v) = cos t; dv. XIII d (cos t^) = — sin v dv. XIV * c2 (tan v) = sec* v rfv, etc. XIX c2(arc sin v) = =y etc. The term differentiation also includes the operation of finding differentials. In finding differentials, the easiest way is to find the derivative as usual, and then multiply the result by dop. Ex. 1. Find the differential of x + 3 ^ xa + 3 _ (itg + 8)dx~(g + 3)2zda; _ {S-6x-z^)dz (x^ + 3)a "" (x» + 3)a ' "** Ex. 2. Find dy from iSoZutton. 2 b^xdx - 2 a^dy = 0. ,\ dy = -—dx. An», Ex. 3. Find dp from ffl:=za^ cos 2 0, Solutum, 2 pdp = - a^sin 2 • 2 d0. a2 8in2^,^ .«. dp=: de. P Ex. 4. Find d [arc sin {St -4 <•)]. SchUum, d [arc am (3 1 — 4 <»)] = ' = - . ^n«. Vl-(3<-4e«)2 Vl - fi 146 DIFFERENTIAL CALCULUS 105. Successive differentials. As the differential of a function is in general also a function of the independent variable, we may deal with its differentiaL Consider the function d(df/) is called the second differential of y (or of the function) and is denoted by the symbol dY Similarly the third differential of y^ {i[(i(dfy)] is written and so on, to the nth differential of y^ dry. Since dx, the differential of the independent variable, is inde- pendent of X (see footnote, p. 140), it must be treated as a constant when differentiating with respect to x. Bearing this in mind we get very simple relations between successive differentials and successive derivatives. For dy=f{x)dx, and dry^r{x){dx)\ since dx is regarded as a constant. Also, d'y=f"{x)(dx)\ and in general d*y =/^">(a;) {dx)\ Dividing both sides of each expression by the power of dx occurring on the right, we get our ordinary derivative notation dj^^^ ^''^' dj^'^ <'^^'"' d^r"'^ ^''^' Ex. 1. Find the third differential of SoltUUm. dy = (6aj* - 6x« + 8)dx, cPy = (20a:«-12x)((to)2, d»y = (60 x» - 12) {dx)*, Ans. Note. This is evidently the derivative of the function multiplied by the cube of the differential of the independent variable. Dividing through by (dx)", we get the third derivative rL?^ = 00x2-12. dx* DIFFERENTIALS 147 DifEerentiate the following, using diflerentials. 1. y = ax«-ea« + cx + <l. dy = (8 ax« - 2 6x + c) das. 2. y = 2a:*-8x'+6ari + 6. dy = (6«* - 2«'"* -6ar«)dx. 8. y = (a2-x?)». dy = -10aj(a»-x«)*cte, 4. y = Vl-fx«. dy = -r=L=dx. Vl-f «« (TT^* ^"(l + x«)-+ SaS^dx 6. y = logVl-x». <'y = 2(xTi,i) ' 7. y = (c* + c-*)». dy = 2(^*-e-««)d«- 8. y = c*logx. dy = (?'(logx + -)dx. l + 8in0 10. p = tan0 + 8ec^. ^""^cofi^S" 11. r = itan»^ + tan^. dr = sec*M^. ^, . , 8(logx)«dx 12. fix) = (log x)». f(z)dx = -^-^ 13. »(Q= . 0'(Od« = (1 - «»)« (1 - «^* dy 16. d[arctanlogy]=: ydy 16. d fr arc vers ?^ - V2 ry - y^l = —=0= L r J V 2 ry — y* 17 dr-^5^-iiogtiui^i=--^- CHAPTER XII RATES 106. The derivative considered as the ratio of two rates. Let y =/(^) be the equation of a curve generated by a moving point P. Its coordinates X and y may then be considered as functions of the time, as explained "Y in § 84, p. 104. Differentiating with respect to ^, by IIYI, we have At any instant the time rate of change of y (or the function) equals its derivative multiplied by the time rate of change of the independent variable. Or, write (81) in the form dp dt The derivative measures the ratio of the time rate of change of y to that of X, — being the time rate of change of length of arc, we have from (12), p. 105, WH^) which is the relation indicated by the above figure. Ex. 1. A point moves on the parabola 6 ^ = x^ in such a way that when x = 6 the abecissa is increasing at the rate of 2 ft. per second. At what rates are the ordinate and length of arc increasing at the same instant ? 148 RATES 149 Solution. Differentiating with respect to the time t, we get dy 1 dx — = - . X — ' dt Z dt This means that at any point on the parabola (Rate of change of y) = (J z) {Rate of change of x). dx But from the problem, when x = 6, — = 2 ft. per sec. dt Substituting in (A), ^ = - . 0.2 ft. per sec. = 4 ft. per sec. dt 3 P(M) Also, — = V4 + 16 = 2 Vs ft. per sec. dt That is, at the point P (6, 6) the ordinate changes in value twice as rapidly as the abscissa. dtf If we consider the point P' (— 6, 6) instead, the result is -^ = — 4 ft. per sec. dt The minus sign indicates that the ordinate is decreasing as the abscissa increases. Ex. 2. A man is walking at the rate of 5 miles per hour towards the foot of a tower 60 ft. high standing on a horizontal plane. At what rate is he approaching the top (a) at any instant ; (b) when he is 80 ft. from -f the foot of the tower ? I Solution, Let y = distance from top and x = dis- i. tance from foot of tower at any instant. Then f x* + (60)2 = y2. { Differentiating with respect to the time t gives *._ dy dt xdx ydi (a) Hence he is approaching the top - times as fast as he is approaching the foot. dx (b) When x = 80, y = 100 ; and we have given — = 5 x 5280. Therefore dt • dt/ 80 -^ = — X 5 X 5280 ft. per hour d« 100 *^ = 4 miles per hour. • Ex. 3. A circular plate of metal expands by heat so that its radius increases uniformly at the rate of .01 inch per second. At what rate is the surface increasing (a) at any instant ; (b) when the radius is 2 inches ? Solution. Let x = radius and y = area of the plate. Then y = irx^, and diflerentiating with respect to the time t, dy a dx dt dt (a) That is, at any instant the area of the plate is increasing in square inches 2 tx times as fast as the radius is increasing in linear inches. 150 DIFFERENTIAL CALCULUS dx (b) When — = .01 in. per sec. and x = 2 in. dt dy -p = .04 r sq. in. per sec. ; i.e. the area is increasing .04 r sq. in. per sec. at that instant Ex. 4. An arc light is hung 12 ft. directly above a straight horizontal walk on which a man 5 ft. in height is walking. How fast Is the man^s shadow lengthening when he is walking away from the light at the rate of 108 ft per minute ? £, Solution. Let x = distance of man from a point directly under light X, and y = length of man*s shadow. From the figure, y : y + « : : 5 : 12, or, y = fx. Differentiating, -r^ = r -r- ; *' dt 7 dt Le. the shadow is lengthening f as fast as the man is walking, or 120 ft per minute. 1. In a parabola y> = 12 x, if z Increases uniformly at the rate of 2 in. per second, at what rate is y increajsing when x = 3 in. ? Ans, 2 in. per sec. 2. At what point on the parabola of the last example do the abscissa and ordi- nate increase at the same rate ? Ans. (8, 6). 3. In the function y = 2 x' + 6, what is the value of x at the point where y increases 24 times as fast as x ? Ans. x = ^ 2. 4. Find the value of x when the function 2 x' ~ 4 is decreasing 5 times as rapidly as X increases. Ara. x = ~ }. 6. Find the values of x at the points where the rate of change of x»-12x2 + 46x-13 is zero. Ans, x = 8 and 6. 6. What is the value of x at the point where x* ~ GX^ + 17 x and x* — Sx change at the same rate ? Ana. x = 2. 7. At what point on the ellipse 16x> + Oy' = 400 does y decrease at the same rate that x increases ? Ans, (3, Y)* 8. Given y = x> — 6x3 + 8x + 5; find the points at which the rate of change of the ordinate is equal to the rate of change of the slope of the tangent to the curve. Ans, X a 1 and 5. 9. Where in the first quadrant does the arc increase twice as fast as the sine ? Ans. At 60^. RATES ISl 10. The side of an equJlatenl txiang^e is 84 inches long, and is increasing «t the rate of 3 inches per hoar ; how £ul is the area increasing ? Ans. 36 Vs sq. in« per hoar. 11. Find the rate of change of the area of a aqnare when the side b is increas- ing at the rate of a units per second. Ans. 2 od sq. units per sec. 12. (a) The volume of a spherical soap babble increases how many times as fast as the radius? (b) When its radius is 4 in. and increasing at the rate of f in. per second, how fast is the volume increasing ? « Ans, (a) 4 «t^ times as fast ; (b) 8S w cu. in. per sec 13. One end of a ladder 50 ft. long is leaning against a perpendicular wall stand- ing on a horizontal plane. Supposing the foot of the ladder be pulled away from the wall at the rate of 3 ft per minute ; (a) how fast is the top of the ladder descending when the foot ia 14 ft. from the wall ? (b) when will the top and bottom of the ladder move at the same rate ? (c) when is the top of the ladder descending at the rate of 4 ft. per minute ? Ana, (a) | ft. per min. ; (b) when 26 V2 ft from wall; (c) when 40 ft. from wall. 14. A barge whose deck is 12 ft. below the level of a dock is drawn up to it by means of a cable attached to a ring in the floor of the dock, the cable being hauled in by a windlass on deck at the rate of 8 ft per minute. How fast is the barge moving towards the dock when 16 ft away ? Ana, 10 ft. per minute. 16. An elevated car is 40 ft. immediately above a surfi^e car, their tracks inter- secting at right angles. If the speed of the elevated car is 16 miles per hour and of the surface car 8 miles per hour, at what rate are the cars separating 6 minutes after they met ? An», 17.8 miles per hour. 16. One ship was sailing sotith at the rate of 6 mUes per hour ; another east at the rate of 8 miles per hour. At 4 p.m. the second crossed the track of the first where the first was two hours before, (a) How was the distance between the ships changing at 3 p.m. ? (b) how at 6 p.m. ? (c) when was the distance between them not changing ? Ana. (a) Diminishing 2.8 miles per hour ; (b) increasing 8.73 miles per hour; . (c) 3.17 P.M. 17. Assuming the volume of the wood in a tree to be proportional to the cube of its diameter, and that the latter increases uniformly year by year when growing, show that the rate of growth when the diameter ia 3 ft is 86 times as great as when the diameter ia 6 inches. CHAPTER XIII CHANGE OF VARIABLE 107. Interchange of dependent and independent variables. It is sometimes desirable to transform an expression involving deriva- tives of y with respect to x into an equivalent expression involving instead derivatives of x with respect to y. Our examples will show that in many cases such a change transforms the given expres- sion into a much simpler one. Or, perhaps x is given as an explicit function of ^ in a problem and it is found more convenient to use - . . . , dx d^x ^ ^, . , . dy d^y ^ a formula mvolvmg -—, ^-r, etc., than one mvolving -^i -— ^, etc. ay djf ax cwr We shall now proceed to find the formulas necessary for making such transformations. Given y:=^f(^^ then from XXYII we have (88) ^ = JL, — :?fcO d,^ d'X cly dy giving -f- in terms of --• Also, by XIYI, ax dy (^) ^ = i-/^\ = A/^\^, or, da? dx\dxj dy\dxj dx do? dy\dx\ dx \dy] \dy) dy Substituting these in {A)y we get (84) p.^^^ ^dy) 152 CHANGE OF VARIABLE 163 giving -z^ in terms of — and — • Similarly, ^ ' dx^ /dx\* • \dy) and so on for higher derivatives. This transformation is called changing the independent variable from x to y, Ex. 1. Change the independent variable from x to y in the equation Schdion, Substituting from (33), (34), (36), dv \dv* a much simpler equation. dy* dy* 106. Change of the dependent variable. Let {A) y=f{^)> and, suppose that at the same time y is a function of Zj say We may then express -^r-^ -7^» etc., in terms of -;-, — -r, etc., as - „ ax dxr dx dor follows. In general, ^e is a function of y by (-B), § 55, p. 57, and since y is a function of x by {A)^ it is evident that z is a function of x. Hence by IIVI we have 154 DIFFERENTIAL CALCULUS Similarly for higher derivatiyes. This transformation is called changing the dependent variable from y to z^ the independent vari- able remaining x throughout. We will now illustrate this process ' by means of an example. Ex. 1. Having given the equation change the dependent variable from y to 2 by means of the relation (F) y = tan «. , Schition, From (F), -^ = 8ec*2-— I -4 = sec«z— - + 28ec»«tan«i — l . ' dx dx dx^ dx^ \dx/ Substituting in (E), a ^2^ . o 4^* /<^\* -i . 2(l + tanz)/ - dz\* sec^z— - + 28ec*2;tana?i — ) =1+ — r^(8ec««~l, dx« \dx/ l + tan«« \ dxJ andreducing, weget— - — 2( -— ) =co8*«. Ana, dx^ \dx/ 109. Change of the independent variable. Let y be a function of x^ and at the same time let x (and hence also g) be a function of a new variable t It is required to express dv d^y ax dor in terms of new derivatives having t as the independent variable. ^ * dy __dg dx dt dx dt dy (A) ^ = ^. ^ ' , doc dM_ at ±/dy\ d^y _d /dy\ _ d /dy\ dt dt \dx) d7? dx\dx) dt\dx) dx dx di But differentiating (A) with respect to ^ dt\dx) dt ldy\ dxd^y dy d^x dt dx \Ttj dt df dt df CHANGE OF VARIABLE 166 Therefore {B) dap d*y dy dhp d^ _ dt dt* dt d^ , \dt) and so on for higher deriyatives. This transformation is called changing the independent variable from x to t It is usually better to work out examples by the methods illustrated above rather than by using the formulas deduced. Ex. 1. Change the independent variable from x to £ in the equation "^S-^^l--"' by means of the relation Solution. ' (S) Also, dx -— = e*. therefore dt ' dt dz — 0-t W dy dy dt ^, ^ -7^ = -^ — - , therefore dx dt dz ^ = er'^. Also, dx dt dx* dxKdt/ dt dx dt\dt/dx dt dx Substituting in the last result from (E), dafi di^ dt ' Substituting (D), (F), (€F) in (C), = 0; and reducing, we get (2*1/ ^ + y = 0. Ans. Since the formulas deduced in the Differential Calculus gener- ally involve derivatives of g with respect to x^ such formulas as (A) and (B) are especially useful when the parametric equations of a curve are given. Such examples were given on pp. 92-97, and many others will be employed in what follows. 156 DIFFERENTIAL CALCULUS 110. Simultaneous change of both independent and dependent variables. It is often desirable to change both variables simul- taneously. An important case is that arising in the transfoimar tion from rectangular to polar coordinates. Since x=:p cos and y=^p sin 0, the equation becomes by substitution an equation between p and 6 defining p as a function of 0. Hence /?, 2:, y are all functions of 0. Ex. 1. Transform the formula for the radius of curvature (40), p. 103, [■-(l)T W 2J = into polar coordinates. Solution, Since in (A) and (£), pp. 164, 155, t is any variable on which x and y depend, we may in this case let t = $, giving dy (5) {C) ox ox dz d*y _ dy cPx d^y _ ded^~^d^ \de) Substituting {B) and (C) in (A), we get \de) \dB/ ded0^ ddde» 2J = ( dx\a d$) /dx\« Kde) ■» or. (D) U = dx d^y _^ dy d?x dSd^"" dSd^ But since x = p cos 9 and y = psin 6, we have — = — psm^ + cos^-f-; -^ = pcoB^ + 8in^~; d^ dtf d^ '^ dtf' ^ = -pcos^-2sin^3^ + cos^^; ^ = - psin^ + 2co8^^ + sin^^. d^ dS d0^ de^ d$ dff^ CHANGE OF VARIABLE 167 Substitating these in (D) and reducing, B= — — -. ^iw. '^ KdeJ *^dfi EXAMPLES Change the independent variable from x to y in the four following equations. dx\«n* dy. Mtn' . . [■+(!)*] 1. J8 = Ara. i? = - dx« dya dx« Vdx/ dy« dy 8. X— ^ + ( ~ ) - T^ = 0. Am. x^-:-H-(t-) =<>• dx« \dx/ dx dy'^ \dy/ \ dx )\djfi) \ dx Jdxdsfi Kdy^J \dy Jdy* Change the dependent yariable from y to z in the following equation. Change the independent yariable in the following eight equations. dx« 1 - x2 dx 1 - x« • dt^ '' 7. (i_x«)^-x^ = 0, x = co8z. Am. ^ = 0. ^ ^dx^ dz dz^ 8. (l-y»)??-y?^ + o*M = 0, y = 8inx. ^n«. ^ + a^u = 0. dy* dy dx* dx* dx X* « dz* d*p -. - d*u du ^ d'v 10. x»^ + 3x2^ + x5^ + r = 0, x = c«. Am. ^ + v = 0. dafi dx^ dx d^ 11. ^ + .lfL*^ + _y_ = o, x = tan^. ^n«. ^ + y = 0. dx* 1 + X* dx (1 + x2)2 * de^ ^ 12. h «u h sec* « = 0, « = arc tan t "^^ *• d*u du Am. (l + <*)^ + (2« + MarctanO — + 1 = 0. di* df 13. x«?| + a«v = 0. » = 1. 4«.. ^-?^ + «ay = o. dx* z dz* z dz 168 DIFFERENTIAL CALCULUS In the following four examples the eqaationa are given in parametric form. dv d^y Find -^ and —4 in each case. dx dx^ 14. X = 7 + 1«, y = 3 + «a - 8t*. Arut, ^ = l-6««, ^ = -6. dx dx^ 15. « = cote, y = 8in»«. Am. ^ = -3Bin*tco8e, !^ = 38in««(4 - 68in«t). dx dx* 16. z = a(co8 1 + < sin t), y = a(8in t — t cos t). Ana, -^ = tan t, dx dafi atcoe^t 17. x = ; — -» y = - — -• Ans, —- = — 1, --— = 0. l + tl + t dxdx* dy 18. Transform — - by assuming x = p cos 6^ y = pmn0. Ans, 19. Let /(x, y) = be the equation of a curve. Find an expression for its slope ( -^ ) in terms of polar codrdinates. ^ . , ^dp \dx/ ^ pcos^ + sm^^ A dy d$ — psm^ + costf-?- '^ d$ CHAPTER XIV CURVATURE. RADIUS OF CURVATURE 111. Curvature. The shape of a curve depends very largely upon the rate at which the direction of the tangent changes as the point of contact describes the curve. This rate of change of direction is called curvature and is denoted by K. We now proceed to find its analytical expression, first for the simple case of the circle, and then for curves in general. 112. Curvature of a circle. Consider a circle of radius E. Let r = angle that the tangent at P makes with OX, and r + At = angle made by the tangent at a neighboring point F'. Then we say At = total curvature of arc FF', If the point F with its tangent be supposed to move along the curve to P', the total curvature (= At) would measure the total change in direction, or rotation, of the tangent; or, what is the same thing, the total change in direction of the arc itself. Denoting by s the length of the arc of the curve measured from some fixed point (as A) to P, and by As the length of the arc FF^^ then the ratio At A« measures the average change in direction per unit length of arc.* Since from the figure. or, As=^B' At, A« b' * Thus, If At s. - radlanf (« a(n» ftnd a«b 3 oentimeten ; then At » ^ — s — radians per centimeter*- 10" per centimeter >- aTerage rate of change of direction. A« 18 169 160 DIFFERENTIAL CALCULUS it is evident that this ratio is constant everywhere on the circle. This ratio is by definition the curvature of the circle^ and we have (86) The curvature of a circle equah the reciprocal of its radiu9. 113. Curvature at a point. Consider any curve. As in the last section, At = total curvature of the arc FP\ At = average curvature of the arc FF\ and A^ More important, however, than the notion of the average curva- ture of an arc is that of curvature at a point This is obtained as follows. Imagine P' to approach P along the curve, then the limiting value of the average curvature ( = — j as P' • approaches P along the curve is defined as the curvature at P, i.e. Curvature at a point = . ^ ( -— ) =-r-' ^ A« = \A8j d% (87) X = -—- = curvature. ds If we suppose P to move along the curve, t and s are functions of the time U and we may write dr j^_dT __dt dt dr where 3- = angular velocity of rotation of the tangent, dt and — = V = magnitude of the velocity of P in the path. Let t; = 1, then J5r= dr dt and we have the theorem : The curvature at P is equal to the angular velocity of the tangent at P when P describes the curve with unit velocity. CURVATURE 161 114. Formulas for curvature. It is evident that if in the last section, instead of measuring the angles which the tangents made with OX, we had denoted by t and t + At the angles made by the tangents with any arbitrarily fixed line, the different steps would in no wise have been changed, and consequently the results are entirely independent of the system of coordinates used. How- ever, since the equations of the curves we shall consider are all given in either rectangular or polar coordinates, it is necessary to deduce formulas for K in terms of both. We have dy or, T = arc tan-p- ax Differentiating with respect to «, ^(dy\ dr ds \dxj tanT = ^, §45, p. 44 Therefore, substituting in (A), dr d^ and, since jr = -;-i we have d% (88) iST = '• ['+(i)T by in " Hi) d^y dx dr da? ds ^ (A) ^= ^Tva ' ByllYI 1 + B.t g4= :■..,. ■ By(«),p,142 S 11 + 162 DIFFERENTIAL CALCULUS If the equation of the curve be given in polar coordinates, K may be found by transforming* the above formula by means of the relations x^p cos d, y=zpsm 01 giving (S9) X=- ^ ^^^ \f^{%n de Ex. 1. Find the curvature of the parabola ffi = ^pz at the upper end of the latus rectum. Solution. ^^2p.d«y^_2£^^_4^ dz y dx^ y* dx y* Substituting in (38), ^ , (y2 + 4 p2)* ' giving the curvature at any point. At the upper end of the latus rectum (p, 2p) (4p« + 4pa)* 16 V^p* 4 V2p Ex. 2. Find the curvature of the logarithmic spiral p^e^ &i any point SohUian. ^^a^ = ap; -^ = alki^ = aV d$ ^' d^ Substituting in (39), 115. Radius of curvature. By analogy with the circle [see (86) p. 160], the radius of curvature of a curve at a point is defined as the reciprocal of the curvature of the curve at that point. Denot- ing the radius of curvature by By we have <> The details of this tranBformatlon were given in Ex. 1 on pp. 156, 167. t While in our work it is generally only the numerical value of K that is of importanoe, yet we can give a geometric metuiing to its sign. Throughout our work we have taken the positive sign of the radical v 1 + (-p) • Therefore A" will be positive or negative at the same time as -j^ 1 that is (§ 98t p. 136), according as the curve is concave upwards or concave downwards. t Hence the radius of curvature will have the same sign as the curvature, i.e. 4- or -> according as the curve is concave upwards or concave downwards. CURVATURE 168 or, substituting the values of K from (38) and (89), [> + (i)T (40) JB = .= ['■^('^)T '•-'^+KI)" a « X Ex. 1. Find the radius of curvature at any point of the catenary y = - (e» + e <>) . Soitition. ^ = 2 («"-«"*) ; ^ = o~ (^ + ^"")- Substituting in (40), XX XX 4. a * e°H- e « e» 4- e « 2a 2a If the equation of the curve is given in parametric form, find the first and second derivatives of y with respect to x from {A) and {B)y p. 154, 155, namely: dy dt dx d^y dy d^x dJ^y 'dide~'did^ ^ \dtj and then substitute the results in (40). Ex. 2. Find the radius of curvature x)f the cycloid X = a (« — sin t), y = a(l — cosQ. Solution. -— = a(l — cost), — = asint; dt dt — - = asmt, -4 = a cost <^ <ia. 164 DIFFERENTIAL CALCULUS Substituting in (B) and (0), and then in (40), p. 163, we get dy smt cPy a (1 — cost) a cos t — a sin t a sin t _ 1 5x~l -cost' dx^" a«(l-cost)» " a(l-co8t)« ^^L Vl-cont/J ^_2aV2_2coBf. 4m. a a(l-co8 0« Note. From (6), p. 90, we get length of nonnal = y Vl + (^J = «(1 - «08 «) Vl + (i3^^ =a V2 — 2 cos f . Hence from a comparison of the last two results : At any point on the cycloid the length of the radius of curvature is twice the length of the normal. EXAMPLES 1. Find the radius of curvature of the equilateral hyperbola xy = 12 at the point (3, 4). An8, R = ^. 2. What is the curvature of y = x* — 4 a^ — 18 x' at the origin ? An8. K= -Z6, 3. Find the radius of curvature of the ellipse iW -f a^ = aV^ (a) at any point, (b) at end of major axis, (c) at end of minor axis (a > h), o*o* a h 4. What is the radius of curvature of the curve 16^^ = 4x* — x* (a) at (0, 1), (b) at (2, 0) ? An^ (a) iJ = 0, (b) iJ = 2. 6. Find the curvature of the cubical parabola a^y = x*. _ . - _, oa*x Ana. K = (a*-f 9x*)* 6. Get the radius of curvature in the semicubical parabola ay^ = z\ A^, ^^x*(4a + 9x)« 6a 7. Find the radius of curvature of the curve y = x* + 6 x* -f- 6 x at the origin. Ana, R = 22.506. 8. Find the point on the parabola y^ = 8 x at which the radius of curvature isT^J. Ana. (}, 8). 9. Find the curvature of the curvef - ) + ( - ) = 1 at the point (0, b). \a/ \b/ 86 Ana. K = — CURVATURE 166 10. Determine the radius of curvature of the curve a^ = bx^ -f- ca^ at the origin. Aii», R = co. 11. Shove that the radius of curvature of the witch y* = — at the vertex 1 ^ ••i- 12. Find the radius of curvature of the curve y = log sec x at any point. Atis, 12 = sec X 13. Find K at any point on the parabola x* -f y* = a*. Ans, K = 2(x + y)' 14. Find R at any point on the hypocycloid x^ + y' = a'. Aii^. 12 = 3 {axy)^ 16. Find R at any point on the cycloid x = r arc vers - — V2 ry — y*. .4ns. 12 = 2 V2ry Find the radius of curvature of the following curves at any point. 16. The circle p = a sin B, Yt. The spiral of Archimedes p = oB. 18. The cardioid p = a (1 - cos ^ 19. Thelemniscatep> = a3cos29. $ 20. The parabola p = a sec' - . $ 21. The curve p = a sin* - . 22. The trisectrizp = 2acos0-a. 23. The equilateral hyperbola p^cos 2 tf = a*. An8. R = - 2 2\l Ans, 12 = (e±^ p2 + 2a« Ana. 12 = fV2ap ^ns. 12 = — 3p $ Ans, JS = 2a8ec'- 2 Ans, 12 = }asin3~ ^ns. j^^^(5-4cos^)t 9 - 6 cos 9 .4ns. 12 = ^ a2 ox on. 1 a(l-c2) . -, a(l-e«)(l-2ccos^-f c*)* 24. The conic p = —^ -• Ans, 12 = -^ f-^ ^ ^ 1 — e cos ^ (1 — c cos ^)' 26. The curve l*~of' « 26. The hypocycloid j^~* X = acos'f, Sinn. Ans, 12 = 4(l+<2)2 Ans, 12 = 3 a sin t cos t, 27. The curve 2a The curve Ans, R = at X = a (cost + tsinQ, y = a (sin t — t cos Q. Sx = a (m cos i + cos mt), m-l \ 2 / CHAPTER XV ^b,o) THEOREM OF MEAN VALUE. INDETERMINATE FORMS 116. Rolle's Theorem. Let y=f{x) be a continuous single-valued function of x vanishing f or x = a and a; = 6, and suppose that /'(x) changes continuously p when X varies from a to 6. The function will then be represented graphically by a contin- uous curve with a con- tinuously turning tan- gent as in the figure. Geometric intuition shows us at once that for at lecutt one value of x between a and b the tangent is parallel to the axis of X (as at P), i.e. the slope [=f'{x)] is zero. This illustrates Rolle's Theorem : Iff{x) vanishes when x = a and a; = 6, and y f{x) andf{x) are continuous for all values of x from x=za to x==b^ thenf{x) will be zero for at least one value of x between a and b,* This theorem is obviously true, because as X increases from a to 6, f{x) cannot always increase or always decrease as x increases, since f{a) = and f(b) = 0. Hence for at least one value of x between a and 6, f{x) must cease to increase and begin to decrease, or else cease to decrease and begin to increase ; and for that particular value of X the first derivative must be zero (§ 93, p. 118). * The second figure shows the graph of a function which is discontinuous (» w ) for x ■» c, a value lying between a and 6. The third figure shows the graph of a continuous function whose first derivatiye does not exist for such an intermediate value x = c. In each case it Is seen that at no point on the graph between x = a and x=b does the tangent (or curve) become parallel to OX, 106 THEOREM OF MEAN VALUE 167 117. The Theorem of Mean Value.* Consider the quantity Q defined by the equation or- ating ((7) we get Therefore, since Let F(x) be a function formed by replacing & by a; in. the left- hand member of (B); that is, ( C) F{x) =f(x) -f{a) - (a: - a) Q. From (5), F(b)=0, and from ((7), F{a)= 0, therefore by RoUe's Theorem, p. 166, F\x) f must be zero for at least one value of x between a and 6, say x^. But by differenti- F^{x)==f{x)^Q. F\x^ = 0, then also/'(a;i) - © = 0, and Substituting this value of Q in (^), we get the Theorem of Mean Valuei (42) fm-f{a) ^ ^,^^^^^ a<x,<b o — a where in general all we know about Xy^ is that it lies between a and b. The Theorem of Mean Value can be easily interpreted geomet- rically. Let the curve in the figure be the locus of Take OC^a and OD=b; then f(a) = CA and f{b) = DB, giving AF=b-^a and FB =f{b) -f{a). Therefore the slope of the chord AB is ^ ^ f(b)^f(a) ^ AE b-a and the Theorem of Mean Value simply asserts that there is at least one point on the curve between A and B (as F) where the tangent (or curve) is parallel to the chord AB. If the abscissa of P is ar^, the slope at P is /'(Xj), and ^« ^^« f(b)-f(a) _ .,. . l-a ~^ ^'>' -/W-/(«) Xx •Also called the Law qf the Mean. t If Fix) and F\x) are oontinuons. 168 DIFFERENTIAL CALCULUS The student should draw curves to show that there may be more than one such point in the interval, and curves to illustrate on the other hand that the theorem may not be true iif(x) becomes discon- tinuous for any value of x between a and 6, or iif'(x) ceases to exist. Clearing (42) of fractions, we may also write the theorem in the form (48) f{h) = f(a) + (6 - «)/'(»!). Let 6 = a -I- Aa ; then b — a = Aa, and since a;^ is a number lying between a and 6, we may write 2;^ = a -f ^ • Aa, where ^ is a positive proper fraction. Substituting in (48), we get another form of the Theorem of Mean Valuei (44) f(a + Aa) - /(a) = Aa/'(a + ff • Aa). < ^ < 1 118. The Extended Theorem of Ifean Value.* Following the method of the last section, let E be defined by the equation (A) fib) -f{a) - (6 - a)f{a) - i (6 - aYE = 0. Let F(x) be a function formed by replacing 6 by a; in the left- hand member of (A); that is, (B) F{x)=f{x)-f{a) - (2; - a)f{a) - J (a: - a^E. From (A), i^(6)=0; and from (fi), F{a)=0, therefore, by RoUe's Theorem, p. 166, at least one value of x be- tween a and 6, say a:^, will cause F'(x) to vanish. Hence, since F'{x)=f'{x)--f(a)-{x-a)E, we get F\x,)=f\x,)^f{a) - {x, > a)E = 0. Since F^ (ar^) = and F^ (a) = 0, it is evident that F' (x) also satis- fies the conditions of RoUe's Theorem, so that its derivative^ namely F*'{x)j must vanish for at least one value of x between a and x^, say a;,, and therefore x^ also lies between a and 6. But F"{x)=f'{x)-E; therefore i^"(a;,)=/''(a;,)- ^ = 0, and E=r{x,). Substituting this result in (A)^ we get ( 0) fib) =f(a) + {h- a)f (a) + ^{h- aff" (x,). a<x,<h •Also called the Extended Law cf the Mean. THEOREM OF MEAN VALUE 169 In the same manner, if we define S by means of the equation f(P) -f(a) - (6 - a)f{a) - i (4 - o)«/"(a) - i (6 _ a)«5 = 0, we can derive the equation {!>) m ==f{a) + (6 - a)f (a) + i (i - a) V" (a) + i (6 - a)-/- (x,), where x^ lies between a and &. By continuing this process we get the general result,* + ^^l~°rV -'>(«) + ^^^>'K). a < ^x < A |n — 1 In where x^.lies between a and b. (E) is called the Extended Theorem of Mean Value. 119. Maxima and minima treated analytically. By making use of the results of the last two sections we can now give a general dis- cussion of maxima and minima of functions of a single independent variable. Given the function f{x). Let A be a positive number as small as we please ; then the definitions given in § 94, p. 118, may be stated as follows: If, for all values of x different from a in the interval [a — A, a + A], (-4) f{x) —f{a) =z= a negative number^ then f{x) is said to be a maximum when x = a. If, on the other hand, (B) f(x) — f{a) = a positive number^ then /(a;) is said to be a minimum when x = a. Consider the following cases: I. Letf(a)^0. From (48), p. 168, replacing 6 by a; and transposing /(a), ( 0) fix) ^f(a) = {x-^ a)f (x,). a<x,<x • It Is aasamed that/(jr), f{x), fix), • • •,/(*) (x) exist throughont the interval [a, b]. 170 DIFFERENTIAL CALCULUS Since /'(«) =^ 0, and f{x) is assumed as continuous, h may be chosen so small that f{x) will have the same sign as /'(a) for all values of x in the interval [a — h,a-\- A]. Therefore /'(a^J has the same sign as /'(a) (§§ 29-33). But x-^a changes sign according as a; is less or greater than a. Therefore, from ((7), the difference will also change sign, and by (A) and (£), /(a) will be neither a maximum nor a minimum. This result agrees with the discussion in § 94, where it was shown that /or all values of x for which f(x) is a maximum or a minimum the first derivative f (x) mu^t vanish. II. Letf{a)=0, and f (a) 4- 0. From ((7), p. 168, replacing J by x and transposing /(a), Since /"(a) i^ 0, and/"(z) is assumed as continuous, we may choose our interval [a — A, a + A] so small that/"(Xj) will have the same sign as /"(a) (§§ 29-33). Also {x — a)' does not change sign. Therefore the second member of (i>) will not change sign, and the difference ^, . ^, . will have the same sign for all values of x in the interval [a — A, a + A], and moreover this sign will be the same as the sign of f"{a). It therefore follows horn, our definitions {A) and {B) that {E) f(a) is a maximum iff '(a) = andf'\a) = a negative number; (F) f{a) is a minimum iff'{a) = andf"(a) = a positive number. These conditions are the same as (21) and (22), p. 124. III. Let f\a) =f"{a) = 0, and f\a) ^ 0. From (2>), p. 169, replacing 6 by 2; and transposing /(a), m f{x) -f{a) = 1^ (^ - «) V"(^a). a<x,<x As before, f"'{x^) will have the same sign as /'"(a). But (x — a)' changes its sign from — to + as 2: increases through a. Therefore the difference ^, . >,, . f{x) --f{a) must change sign, and f{a) is neither a maximum nor a minimum. THEOREM OF MEAN VALUE 171 IV. Let f\a) ^f\a) = =/^-'>(a) = 0, and f^''\a) ^ 0. By continuing the process as illustrated in I, II, and III, it is seen that if the first derivative of /(a:) which does not vanish for 2; = a is of even order (= n)^ then (45) /(a) is a maximum if/(>»)(a) = a negrative number; (46) /(a) is a minimum if /(«»)(a) = a positive number.* If the first derivative of /(a:) which does not vanish for a; = a is of odd order, then /(a) will be neither a maximum nor a minimum. Ex. 1. Examine x* — Qx^ + 24x — 7 for maximum and minimum values. Solution, f{z) = x« - 9x« + 24x - 7. /'(x) = 3xa-18x + 24. Solving 8x2 -18x + 24 = gives the critical values x = 2 and x = 4. .•. /(2) = 0, and / (4) = 0. Differentiating again, /"(x) = 6 x - 18. Since /''(2) = - 6, we know from (45) that/(2) = 13 is a maximum. Since /''(4) = + 6, we know from (46) that /(4) = 9 is a minimum. Ex. 2. Examine e' + 2 cosx + e-' for maximum and minimum values. Solution, f{x) = e*-f-2cosx + c-*, /'(x) = e« - 2 sin X - C-* = 0, for X = 0,t jT'lx) = c* - 2 cosx -f e-* = 0, for X = 0, /"'(x) = e»= + 2 sinx - C-* = 0, for X = 0, /i^(x) = e* -f 2 cosx -f e-* = 4, for X = 0. Hence, from (46), /(O) = 4 is a minimum. EXAMPLES Examine the following functions for maximum and minimum values, using method of the last section. 1. 8x* — 4 X* -f 1. An8. X = 1 gives min. = ; X = gives neither. 2. x» - 6x2 -f 12 X -f 48. Ana. x = 2 gives neither. 3. (X - l)2(x + 1)». Am. X = 1 gives min. = ; X = I gives max. ; X = — 1 gives neither. 4. Investigate 7fi — 5x* + 5x' — 1, at x = 1 and x = 3. 5. Investigate x'* - 3x" + 3x -f 7, at x = 1. 6. Show that if the first derivative of f(x) which does not vanish for x = a is of odd order (= n), then /(x) is an increasing or decreasing function when x = a, according as/<*)(a) is positive or negative. * As in §94, a critical ralae x=aiB found by placing the first derivatlye equal to zero and flolTing the resaltlng equation for real roots. t X "» is the only root of the equation e«- 2 sin x - e- •— 0. 172 DIFFERENTIAL CALCULUS 120. The Generalized Theorem of Mean Value. This theorem asserts about the two functions f{x) and F{x) that where x^ lies in the interval [a, 6] and F'(x) does not vanish in the interval. Proof. By multiplying both sides of (47) by F'{x^) and trans- posing /'(arj to the left-hand side, we see that this theorem requires that the equation F{b) - F{a) ^ ^ -^ ^ ^ shall have a root x^ lying between a and (. In order to make it possible to apply RoUe's Theorem, p. 166, let us try to construct a function having this left-hand member for a derivative and such that it (the function) vanishes for a; = a and x=b. Such a func- tion is /|5|^[i^(a:)-i^(a)]- [/(.)-/(«)], because it vanishes for x:=a and 2; =? 5, and hence by RoUe's Theorem its derivative must vanish for an intermediate value of 2;, say x^i that is, F{b) - F{a) ^ ^""'^ ^ ^""'^ "' which is equivalent to (47). 121. Indeterminate forms. When, for a particular value of the independent variable, a function takes on one of the forms — 1 ^1 0-QO, QO — 00, 0% 00**, 1*, it is said to be indeterminate^ and the function is not defined for that value of the independent variable by the given analytical expression. For example, suppose we have /(^) ^(^) where for some value of the variable, as 2; = a, f(a) = 0, F(a) = 0. INDETERMINATE FORMS 173 For this value of x our function is not defined and we may therefore assign to it any value we please. It is evident from what has gone before (Case II, p. 28) that it is desirable to assign to the function a value that will make it continuous when 2; = a, whenever it is possible to do so. 122. Evaluation of a function taking on an indeterminate form. If when 2; = a the function /(s;) assumes an indeterminate form, then M takm, as the value off{x) for x==a. The assumption of this limiting value makes f(x) continuous for x=z a. This agrees with the theorem under Case II, p. 23, and also with our practice in Chapter IV, where several functions assuming the indeterminate form ^ were evaluated. Thus, for 2:= 2, the function — assumes the form -» but x — 2 limit a:* — 4 = 4. . a:=2 a:-2 Hence 4 is taken as the value of the function for x = 2. Let us now illustrate graphically the fact that if we assume 4 as the value of the function for 2; = 2, then the function is continuous for 2; = 2. 2:^-4 Let y = 2'-2 This equation may also be written in the form y(2:-2) = (x-2)(2: + 2), or, (2;-2)(y-2:-2)=0. Placing each factor separately equal to zero, we have 2; = 2, and y z= 2; -{- 2. In plotting, the loci of these equations are found to be the two lines AB and CD respectively. Since there are infinitely many points on the line AB having the abscissa 2, it is clear that when 2; = 2 (= OM)^ the value of y (or the function) may be taken as any * The calculation of this limiting valae Is called evcUwUing the indeterminate form. 174 DIFFERENTIAL CALCULUS number whatever, but when x is different from 2 it is seen from the graph of the function that the corresponding value of y (or the function) is always found from y = a; -I- 2, the equation of the line CD. Also, on CD, when a; = 2, we get y = jyrp = 4, which we saw was also the limiting value of y (or the function) f or a; = 2 ; and it is evident from geometrical considerations that if we assume 4 as the value of the function for a:= 2, then the function is continuous for 2; = 2. Similarly, several of the examples given in Chapter IV illustrate how the Umiting values of many functions assuming indeterminate forms may be found by employing suitable algebraic or trigonomet. ric transformations, and how in general these limiting values make the corresponding functions continuous at the points in question. The most general methods, however, for evaluating indeterminate forms depend on differentiation. 123. Evaluation of the indeterminate form §• Given a function of the form ^^-^-^ such that /(a) = and Fia) = ; that is, the ^{^) function takes on the indeterminate form - when a is substituted for z. It is then required to find limit f(x) x = a F(x) * Considering the functions f{x) and F{x) the same as in § 120 and replacing 6 by a; in {iX)^ p. 172, we get F{x) F\x^) ' [Siii06/(a)= 0, and ^(a)= 0.] Since x^ lies between x and a, x^ approaches a as its limit when X approaches a, and we have limit f^ ^ limit f(x,) ^ limit f(x) ^, xz=a F{x) x^ = a F\x^ x=a F'{x) ' f*(x) * Assuming that ' does approach a limit as x approaches a. INDETERMINATE FORMS 175 Iif'(z) divided by F\x) does not assume an indeterminate form for a: = a, we may write (i9i\ limit /(a?) _ /^(«) ^^ » = «!?' (05) JP^a)' where f{a) = 0, i^(a) = 0, F\a) ^ 0. Hence Rule fqr evaluating the indeterminate form g* Differentiate the numerator for a new numerator and the denominator for a new denominator,* The value of this new fraction for the assigned valu£^ of the variable will he the limiting value of the original fraction. In case it so happens that f{a) = and F\a) = 0, that is, the first derivatives also vanish for*a;=a, then we still have the indeterminate form 0' and the theorem can be applied anew to the ratio f{x) F\x) ' giving us limit /(^ ^ f\a) x=aF{x) F'\ay When also /"(a) = and F^\a)=: 0, we get in the same manner limit /(^ ^ f'\a) x=aF{x) F'^\a)' and so on. It may be necessary to repeat this process several times. * The student is warned against the very oareleaa but common mistake of differentiating the whole expression as a fraction by VIII. t If as «, the substitution x^- reduces the problem to the eyaluatlon of the limit for z« ; .^__ limit fix) Umit W^ limit w limit /W Therefore the rule holds in this case also. 176 DIFFERENTIAL CALCULUS ^JLtU] _ 1-3 + 2 ' - X + iJ x=i~ 1 - 1 - 1 + 1 "" O' — 1 T? 1 T? 1 * f(^) «*-3x + 2 Ex.1. Evaluate ^ = -j^_^__^ when x = l. SoUUion. m= x»-3x.f2 F(l) x»-x2 /"(I) ^ '3aa-3 -| _ 3-3 _ F'(l) 3x2 - 2x - lJx=i~ 3 - 2 - 1 "" r(l) ^ 6x 1 ^ _6_ ^3 1?'"(1) 6x-2Jx=i 6-2 2' Ex.2. Evalaate^^"^^^^"^''"^^. X = u X — sin X /(O) _e»-e-»-2x F(0) ~ X /^(O) __ C + g-» - 2 -1 _ 14 -1-2 _ 2?^(0) 1-C08X Jx=o~ 1-1 ""O' F"(0) " sinx Jx=o~ ~0" '*•* r"(0) ^ e' + e- n 1 + 1 -F'"(0) cosx indeterminate, indeterminate. Solution, n _ i-i-0 _ Jx=o~ 0-0 Jx=0 1 .*. indeterminate. . indeterminate, indeterminate. Ana, EXAMPLES Evaluate the following by differentiation.* J limit xg-16 x = 4x2 + x-26 2 limit x-1 x = lx»-l' 3 limit logx x = la._i' ^ limit e*-g-* 35 = *i sin X K limit tan X — X o. f. • X = u X — sin X Q limit log sin x ' x = ^(T-2x)a' 7 limit a'-b* ^- x = 0--7~- g limit r» - gf^ - aV + a» ' r = a »^ - a^ Ana. -' 1 n 1. 2. 2. 1 ^ ■ 8 1 A log-. Q limit ^ — arc sin $ * = sin8^ jQ limit sin X- sin X=0 x-0 jj limit ey -f sin y ^ 1 ' y = log(l + y) io limit tan^-f sectf-1 6' — tan ^ - sec ^ + 1 23 limit 8ec20 -- 2tan0 = ^ 1 + cos 4 An^. — . 6 COS0. 2. 1 — I 2 0. l^ limit gz-g^ ^ « = a a* - 2a»z + 2a2» - z*' ^'^ 15. limit J^^e^)!^ x = 2(x^4)e' + e2x • After differentiating, the stadent shonld in every case reduce the resulting expression to its simplest possible form before substituting the value of the variable. INDETERMINATE FOiUVlS 177 124. Evaluation of the indeterminate form §• In order to find ^"^'^ ^ x = a F{x) when i^^*^ fix) = 00 and i^™^^ Fix) = oo, that is, when for x^a the function fix) F{x) assumes the indeterminate form 00 00' we follow the same rule as that given on p. 175 for evaluating the indeterminate form -• Hence Rule for evaluating the indeterminate form ^ • Differentiate the numerator for a new numerator and the denominator for a new denmainator. The value of this new fraction for the assigned value of the variable will be the limiting value of the original fraction,* In case the new fraction is indeterminate for the given value of the variable, we repeat the process as in the last section. To prove this rule we must show that Umit /(£) ^ f'ja) x = a F{x) F\a) when ^"^l fix) = 00, I'^l Fix) = oo. First we assume that 4:77-7 = U where Z is a definite number. x = a F\x) From the Generalized Theorem of Mean Value, p. 172, we have ' F{x)-F(b) F'{x,) 1 ^ [Replacing h and ahj x and 6 respecttvely.] where b is an arbitrary number and F\x) does not vanish in the interval [a, a -f A]. From (A) or, fix) = m + ^ [F{x) - F(b)l * f'{x) and F\x) are aaBomed to be continuoua. 178 DIFFERENTIAL CALCULUS Dividing through by F(x), (^ F{x) F{x) F\x^) L F{x)\ In (£), let X approach a as a limit. Then limit /(^ ^ limit f{x^) x = a F{x) x=ia F\x^ ' Now let h approach the limit zero ; then x^ will approach the limit Oj and we get • limit f(x) ^ f(a) x=aF{x) F\a) In the same manner the rule may be shown to hold true when limit f{x) __ x=^a F\x) oo. Ex. 1. Evaluate -^- for x = 0. C8CX SohdioTL f(0) _\ogx F(0) caox -Ixi-O — CO ■ ■ I CD .'. indeterminate. F'(0) F"(0) 1 X — cscxcotx, 2Binxcosx Ll««o Bin^x XCOB .*. indetezminate. coax — xsinxJx-o ] = — ^ = 0. ^fw. 125. Evaluation of the indeterminate form 0*oo. If a function f{x) • <^ {x) takes on the indeterminate form * oo for rr = a, we write the given function <t>(x) m. 00 so as to cause it to take on one of the forms x or — , thus bringing it under § 123 or § 124. INDETERMINATE FORMS 179 Ex. 1. Evaluate sec 8 x cos 6 x for x = - • 2 SobdUm, sec 8 X cos 5 x]x^ * = oo • 0. /. indeterminate. 1 cos 6 X fisiS Substituting for sec 8x, the function becomes = ^^-^-i-. cob8x co8 8x F{x) f(-) \2/ cos6x-| , , , , ^ — = -— , = - • .-. indeterminate. /T\ cos3xJx«- F r(-) ^ \2/ -sin5x.6-| 6 . ^,(Z.\ -8m3x.8Jx,| 8 126. Evaluation of the indeterminate form oo — oo. It is possible in general to transform the expression into a fraction which will assume either the form - or ^. 00 Ex. 1. Evaluate sec x — tan x for x = - . 2 SaMXoii, sec X — tan xlx»* = oo — oo. .*. indeterminate. By Trigonometry, secx-tanx = -i- - ?ilii^ = Ln^l^ = ZM. * ^ *" COSX COSX C08X F{^) Ai/ 1-sinxi 1-10 , , , , ^ — = . = ^ = - • .-. indeterminate. _/T\ COSX J*-? _,Vir\ — SinxJac-!^ —1 EXAMPLES Evaluate the following expressions by differentiation.* y limit ax' + 6 . a limit tan^ -^^ , 2 limit cotx x = Oiogx 3 limit logx ^ fi' ^^'i j;;7^ ^• * X = 00 x* * In solving the remaining examples in this chapter It may be of assistance to the student to refer to J 37, pp. 36, 36, where many special forms not indeterminaie are evaluated. Ans. -• c 4. limit tan^ ^ = ^Un8^' 2 00. 0. 6. Umit V 2/ = - tan0 2 180 DIFFERENTIAL CALCULUS limit y 6- - ^n*0. 12. Mmur 2 _ 1 1 _1 7. ^^'"^i (X - 2 X) tan X. 2. 13. ^^\ T-l- _ _^1 . ^ i. x = - » = lLlogx logxJ 8. "°^^^x8in2. a. 14. ^^^i [see ^ - tan ^]. 0. X=ao X tf = - 2 9. l^^^^o^-logx. [npcltive.] 0. 15. »™ij. [ 2 _ l_n 1 x = ° = OL8in20 1-co8aJ 2 ,^ limit., * ^v oi. 1 i« limit ry_ 1 1 1 10. ^ = r(l-tan^)8ec2(^. 1. ^^ y = 1 Lj^^ - i^J' a' ' 11. >i°^iMa«-0«)tan^. i^'. 17. "°^^ f^ 1' - ^ = a^" ''' 2a IT « = 0L4« 2«(c»« + l)J 8 127. Evaluation of the indeterminate forms O^i 1", oo^ Given a function of the form In order that the function shall take on one of the above three forms, we must have for a certain value of x /(x)=0, <^(a:)=0, giving 0^ or, f{x) =1, «^ (a;) = 00, giving 1"; or, f{x) = 00, «^ {x) = 0, giving oo^ Let y=f{^y'''\ taking the logarithm of both sides, logy = «^(a;)log/(a:). In any of the above cases the logarithm of y (the function) will take on the indeterminate form Ooo. Evaluating this by the process illustrated in § 125 gives the limit of the logarithm of the function. This being equal to the logarithm of the limit of the function, the limit of the function is known.* • Thus, if Mmit logg y ■= a, then y - ««. INDETERMINATE FORMS 181 Ex. 1. Eyaluate sc* when x = 0. Solution. This function aasames the indeterminate form (fi toTZ = 0. Let y = x*; then log y = X log X = • - By § 126, p. 179, , log X — a X 1 when X = 0. By § 124, p. 177, logy = — ^ = - x = 0, when x = 0. "x2 Since y = x*, this gives loge x' = ; that is, x' = 1. Ans. I Ex. 2. Evaluate (1 + x)' when x = 0. Solution, This function assumes the indeterminate form 1* for x = 0. 1 Let y=:(l + x)^; then log y = - log{l + «) = oo • 0, when x = 0. X By § 126, p. 179, log y = ??iil±^ = ? , whenx = 0. X 1 + X 1 By § 123, p. 174, logy = -J-- = ; = 1, whenx = 0. 1 1 + X a 11 Since y = (1 + «)*, this gives log. (1 + x)* = 1 ; i.e. (1 + x)* = c. Ana. Ex. 8. Evaluate (cot x)»*»* for x = 0. SohUUm. This function assumes the indeterminate form ooP f or x = 0. Let y = (cot x)"*»* ; then log y = sin X log cot x = • oo, when x = 0. By § 126, p. 179, logy = ?2«5?^ = ^, when x = 0. cscx * — csc^x cot X sin x By § 124, p. 177, logy = > = — — = 0, when x = 0. — cscx cot X cos'x Since y = (cot x)«*n*, this gives log, (cot x)*^* = ; i.e. (cot x)«*»' = 1. Ana. 182 DIFFERENTIAL CALCULUS EXAMPLES Evaluate the following ezpressionB by diflerentiation. X = 1 «• i'nc^r" 3. "™^i(sin^>t"«. ^ 2 y = (X>\ ^ yj e 1. 1. &*. 9. J^^ Q (cos m^)»*. 1 e c". 1 e CHAPTER XVI CIRCLE OF CURVATURE. CENTER OF CURVATURE 128. Circle of curvature.* Center of curvature. If a circle be drawn through three points P^, P^, P, on a plane curve, and if P^ and P, be made to approach P^ along the curve as a limiting posi- tion, then the circle will in general approach in magnitude and position a limiting circle called the circle of curvature of the curve at the point P^. The center of this circle is called the center of curvature. Let the equation of the curve be (1) y=/(^); and let rr^, z^, x^ be the abscissas of the points Po, Pj, Pa respectively, (a', yS') the coordinates of the center, and -B' the radius of the circle passing through the three points. Then the equation of the circle is and since the coordinates of the points P^, P^, P, must satisfy this equation, we have ({X, - a'y + (y, - I3y - 22" = 0, (2) .(x,^ar + {i/,-l3r^E^ = 0, Now consider the function of x defined by F{x)={x^ay + (tf - I3y - B'^, in which y has been replaced hyf{x) from (1). Then from equations (2) we get F{x,)=0, P(a:,) = 0, F{x,)=0. F,C«i.1'«> FiCpCi'l^i^ • Sometimes called the oiculating circle* 183 184 DIFFERENTIAL CALCULUS Hence by RoUe's Theorem, p. 166, F^(x) must vanish for at least two values of :r, one lying between x^ and a?^, say rr', and the other lying between x^ and a^j, say x" ; that is, F\x)=0, F'{x')=0, Again, for the same reason, F"(x) must vanish for some value of X between x' and a?'', say x^ ; hence F'\x,)=0. Therefore the elements a\ yS', B' of the circle passing through the points Pq, Pj, Pj must satisfy the three equations F(x,)=0, F'{x')=0, F"{x,)=0. Now let the points P^ and Pj approach P^ as a limiting position ; then Xy^y x^, x\ x'\ x^ will all approach a;,, as a limit, and the elements a, 13, R of the osculating circle are therefore determined by the three equations P(x,)=0, P'(a:o)=0, P"(Xo)=0; or, dropping the subscripts, what is the same thing, {A) (x-a)»-h(y-/3)» = i?, (P) (a: - a) + (y - /S) -/ = 0, differentiating {A). dx ^^^ ^ "^ (^)' "^ ^^ ■" '^^ S " ^' differentiating (P). Solving (P) and ((7) for a: — a and y — yS, we get, {-^^^y dx [_ \dx) J "a;— a = (-») y-/8 MS d^y d7? hence the coordinates of the center of curvature are P S [' + (S)T . . > + (S)" /^v (S-) CIRCLE AND CENTER OF CURVATURE 185 Sulistituting the values of x — a and y — /3 from (J)) in (J.), and solving for B, we get B = ± [l + /^^Yl' \dxj J d7? which is identical with (40), p. 163. Hence Theorem. The radiu9 of the circle of curvature equ(d$ the radius of curvature. From (23), p. 136, we know that at a point of inflection (as Q in the figure) da^ = 0. Therefore, by (88), p. 161, the curvature Jr= 0; and from (40), p. 163, and (Jg?), p. 184, we see that in general a, y3, B increase with- out limit as the second derivative approaches zero. That is, if we suppose P with its tangent to move along the curve to P', at the point of inflection Q the curvature is zero, the rotation of the tangent is momentarily arrested, and as the direction of rotation changes, the center of curvature moves out indefinitely and the radius of curvature becomes infinite. Ex. 1. Find the coordinates of the center of curvature of the parabola 2^ = 4 px corresponding (a) to any point on the curve; (b) to the vertex. Sol^ion. ^ = !£;^ = _1^. dz y dx^ y* (a) Substituting in (E), . y' + 4pa 2p y« ^o, . «« '4jpa" /3 = y- ya 4jpa 4p« Therefore ( 3 a; + 2 p, — =— ) is the center of curva- \ 4pg/ ture corresponding to any point on the curve. (b) (2 Pf 0) is the center of curvature corresponding to the vertex (0, 0). 186 DIFFERENTIAL CALCULUS 129. Center of curvature the limiting position of the intersection of normals at neighboring points. Let the equation of a curve be (A) y =/(^)- The equations of the nonnaLs to the curve at two neighboring points Pj and Pj are* If the normals intersect at C\a\ 13% (?vVi) the coordinates of this point must satisfy both equations, giving -^CaCo'Vo) (B) Now consider the function of x defined by in which y has been replaced hyf{x) from (A), Then equations (B) show that <^(r,)=0, <l>{x,)^0. But then by Rolle's Theorem, p. 166, <^'(a:) must vanish for some value of X between Xq and x^^ say 2/. Therefore a' and yS' are deter- mined by the two equations <l>{x,)=0, <^'(a:')=0. If now Pj approaches Pq as a limiting position, then x* approaches x^, giving <l>{x,)==0, <l>\x,)=0; and C" (a', /3^ will approach as a limiting position the center of • From (2), p. 90, X and Y being the variable coordinates. CIRCLE AliD CENTER OF CURVATURE 187 curvature C (a, 13) corresponding to F^ on the curve. For, if we drop the subscripts and write the last two equations in the form (»-«')+(y-^')f =». it is evident that solving for a* and yS' will give the same results as solving (B) and (C7), p. 184, for a and ;8. Hence Theorem. The center of curvature C corresponding to a point P on a curve is the limiting position of the intersection of the normal to the curve at P with a neighboring normal, 130. Evolutes. The locus of the centers of curvature of a given curve is called the evolute of that curve. Consider the circle of curvature corresponding to a point P on a curve. If P moves along the given curve, we may suppose the correspond- ing circle of curvature to roll along the curve with it, its radius varying so as ^^«55>s^r:::^*"-^^^^ to be always equal to the radius of cur- />cV~'"A, / vature of the curve at the point P. The ( >» K ^ '1 ./o curve CCj described by the center of the V ^\ \\J/Pt circle is the evolute of PP^. ^-O,.*^ Formula (E)^ p. 184, gives the coordi- ■'' nates of any point (a, 13) on the evolute expressed in terms of the coordinates of the corresponding point (a-, y) of the given curve. But y is a function of 2;, therefore ['<g)']^ , '-m a=^z-i= i^ — ^-=i — , /3 = yH ^ — ^ d^ ^ ^ dJ dx" give us at once the parametric equations of the evolute in terms of the parameter x. To find the ordinary rectangular equation of the evolute we eliminate x between the two expressions. No general process of elimination can be given that will apply in all cases, the method 188 DIFFERENTIAL CALCULUS to be adopted depending on the form of the given equation. In a large number of cases, however, the student can find the rectan- gular equation of the evolute by taking the following steps. General directions for finding the evolute. First step. Find a and fifrom (JB), p, 184. Second step. Solve the two resulting equations for x and y in terms of a and 13, Third step. Svhstitvte these values of x and y in the given equa- tion. This gives a relation between the variables a and fi which is the equation of the evolute. Ex. 1. Find the equation of the evolute of the parabola of y^ = 4px. dy 2p d^y 4p2 SolvJLwn, First step. Second step. Third step. dx y dx^ y' o = 3x + 2jp, /3 = - 4p3 p/3a=l(a-2p)«. Remembering that a denotes the abscissa and /3 the ordinate of a rectangular system of co5rdinates, we see that the evolute of the parabola AOB is the semicubical parabola DC'E\ the centers of curvature for O, P, Pi, Pj being at C, C, Ci, C% respectively. Ex. 2. Find the equation of the evolute of the ellipse If^x^ + aV = a^^- Solution. -^ = > — - = dz a^y dx^ oV First step. Second step. *= a* ' (a^-l^)y^ Third step. (oa)* + (hfi)^ = (a* - 62)', the equation of the evolute EUE'B' of the ellipse ABA'R. E, W^ H\ H are the centers of curvature corresponding to the points Ay A\ Bj Bf on the curve, and C, C, C" correspond to the points P, P*, P", CIRCLE AND CENTER OF CURVATURE 189 When the equations of the curve are given in parametric form, we proceed to find -~ and •--^, as on pp. 154, 156, from ax dor (^) {B) dy rfy __ dt dx dx di dx d?y dy d?x ^y Jdt'dj^'"dt'd^ ^ d3? "" y^Y ' and then substitute the results in formulas (27), p. 184. This gives the parametric equations of the evolute in terms of the same parameter that occurs in the given equations. Ex. 3. Find the parametric equations of the evolute of the cycloid. |x = o(<-BinQ, ^ ' |y = a(l — cost). Sohdion, As in Ex. 2, p. 163, we get dy Bxnt d^ 1 dx 1-cost dx* a(l-co8Q« Substituting these results in for- mulas (E), p. 184, we get im (« = «(* + sin <), ^ ' (/3 = -a(l-cos<). Ana. Note. If we eliminate t between equations (2>) there results the rectan- gular equation of the evolute 0(y(f referred to the axes (Xa and (X/3. The co5rdinates of with respect to these axes are (— ira, —2a). Let us transform equations (D) to the new set of axes OX and OY, Then = 0: — ira, /3 = y-2a, « = f-T. Substituting in (D) and reducing, the equations of the evolute become fx = a(f — sinf), y = a (1 — cos t^. Since (J?) and (C7) are identical in form, we have : Tht eoolute of a cycloid is itself a cycloid whose generating circle equals that of the given cycloid. {E) 190 DIFFERENTIAL CALCULUS 131. Properties of the evolute. Differentiating a and 13 from (£7), p. 184, with respect to x gives (^) (^ da _ dy dx \d3?) d:f \^^) ^^ dx dx /^S[^* dl3 dx\d7?) da? \dxj da? Dividing (5) by (A), member for member, ^ ' da di dx JO But -^ = tan t' = slope of tangent to the evolute at Cy and da -^ = tan T = slope of tangent to the given curve at the corresponding point P. Substituting the last two results in ((7), we get tan t' = . tanr Since the slope of one tangent is the negative reciprocal of the slope of the other, they are perpendicular. But a line perpendicular to the tan- gent at P is a normal to the curve. Hence A normal to the given curve is a tangent to its evolute. From (12), p. 105, and (A) and (jB), we have for an arc « of the evolute L \dx) J I /d^jA dx'J CIRCLE AKD CENTER OF CURVATURE 191 But tbia is identically the result we get bj differentiating B, (|0), p. 163, with respect to z and then squaring. Therefore \dxj \dx) ' or, A» = ± an. That is, the radius of curvature of the given curve increoBes or deereageg at fait aa the arc of ike evoltUe increaeet. In our figure this means that p(,^ _ pf,^ ^^^ ^^^ The length of an arc of the evolute ia equal to the difference between the radii of curvature of the given curve which are tangent to thit arc at ita extremities.* Thus in Ex. S, p. 189, we observe that if we fold Q'P' (= 4 a) over to the left on the evolute, F" will reach to 0\ and we have : The length of one arc of the cydoid (at OO'Q^ ia eight timea the length of the radiua of the generating circle. 133. Involutes and their mechanicAl construction. Let a flexible ruler be beat in the form of the curve C,C^ the evolute of the curve P,P,, and suppose a string of length B, with one end fastened at C, to be wrapped around the ruler (or curve). It is clear from the results of the last section that when the string is unwound and kept taut the free end will describe the curveJ*,i*,. Hencethenameeuo?«(e. The curve P,P, is said to be an involute of (7,C,. Obviously any point on the string will describe an involute, so that a given curve has an infinite number of involutes but only one evolute. The involutes P,P„ Pi'^t'i P"Pt' are called parallel curves since the distance between any two of them measured along their com- mon normals is constant. The student should observe how the parabola and ellipse on p. 188 may be constructed in this way from their evolutes. ■ It I* iMDmed that -— <lo«a Dot otaugs ilgn. 192 DIFFERENTIAL CALCULUS EXAMPLES Find the coordinates of the center of curvature and the equation of the evolute of each of the following curves. 1. The hyperbola — — ~ = 1. Ans, a = ^ — - — '- — , /3 = — ^ — — — '-^ - a* cr a* 6* evolute (da)* - (6/3)' = (a^ + 6>)'. 2. The hjrpocycloid x* + y* = a*. Ans. o = x + 3x*y*, /3 = y + 3 x'y* ; evolute (tt + ft* + (o - /3)' = 2 a'. 8. The cycloid x = r arcvers =- — V2 ry - y^. iln«. tt = x + 2 V2 ry — y«, /3 = — y; evolute o = r arcvers ( - -) + V- 2r/3 - /S^. 4. The semicubical parabola x^ = ay^. /v « ' r evolute 720 a/J^ = 16 [2 a + Va^ ~ 18 ao]« [ Va* - 18 oa - a]. 6. The tractrix X = a log ^^^-^^^^ — - Va^ - y^. y a + Va^^r^ a« , a( - --\ Am. a = alog =^» /3 = —; evolute /3 = -\c« + e «/. a - -- V 5 _* 6. The catenary y = -(ef^ ■\- e «). ^n«. o = x — -(c« — c «), /3 = 2y; 2 2 evolute o = a log ^ ^ (^ ""* ^ ) :p A (sa _ 4 ^s)*, 2a 4a 7. Fhid the coordinates of the center of curvature of the cubical parabola y* = a%;. a* + 16y* ^ a*y~9y« 6a2y "^ 2 a* 8. Show that in the parabola x^ + y^ = a* we have the relation a + /9 = 3 (x + y). 9. Find the equation of the evolute of the cissoid y^ = 2a ~x Am. 4006a«a + 1152a«/3a + 27/5* = 0. 10. Given the equation of the equilateral hyperbola 2xy = a^; show that - , o_ (y + g)' ^ o _ (y - g)' From this derive the equation of the evolute (a + /9)' - (a — ft' = 2 a*. Find the parametric equations of the evolutes of the following curves in terms of the parameter t. 12. The curve * = ^f' ^««. « = *<! + 2''-'*). 18. The curve |* = "/^^ +'""«|' Ar^. j » = " f /' (y =a(8int-<cost). (/8 = a8in«. CHAPTER XVII PARTIAL DIFFERENTIATION 133. Continuous functions of two or more independent variables. A function /(a:, y) of two independent variables x and y is defined as continuous for the values (a, b) of {Xy y) when limit 25 = a/(^ y) =/(«i 6)» y = h no matter in what way x and y approach their respective limits a and J. This definition is sometimes roughly summed up in the statement that a very small change in one or both of the independ- ent variables shall produce a very small change in the value of the function.^ We may illustrate this geometrically by considering the surface represented by the equation Consider a fixed point P on the surface where a: = a and y = i. Denote by Aa; and Ay the increments of the independent vari- ables X and y, and by Lz the corresponding increment of the dependent variable 2, the coordinates of P' ^"^ (a; 4. Aa:, y -f Ay, 2 4- A2). At P the value of the function is z =:f(a, b) = MP. If the function is continuous at P, then how- ^ ever Aa; and Ay may approach the limit zero, Az will also approach the limit zero. That is, M'P' will approach coincidence with MP^ the point P' approaching the point P on the surface from any direction whatever. * This will be better imdentood if the student again reads oyer § 33, p. 22, on oontinnoufl functions of a single yariable. 103 194 DIFFERENTIAL CALCULUS A similar definition holds for a continuous function of more than two independent variables. In what follows, only values of the independent variables are considered for which a function is continuous. 134. Partial derivatives. Since x and y are independent in X may be supposed to vary while y remains constant, or the reverse. The derivative of z with respect to x when x varies and y remains constant* is called the 'partial derivative of z with respect to x^ and dz is denoted by the symbol — . We may then write dx ... dz_ limit V f(x + i^y) -fix, y) -| ^^> a;-Aa; = 0L Ai J' Similarly, when x remains constant* and tf varies, the partial derivative of z with, respect to y ia ,Ji\ ?£_ limit r fjr, y + Ay) -f(x, y) ") ^^> 8y-Ay = 0L Ay J* — is also written — f(x^ y) or — ; similarly dx dx*^ ^ ^^ dx '' — is also written — /(ar, y) or — . dy dy' ^ ' ^^ dy In order to avoid confusion the round ^f has been genei*ally adopted to indicate partial differentiation. Other notations, how- Bver, which are in use are (di)' \£) ' ^^^^' y)'-^/(^' y) ' f-^^' y)Jy{^^ y)y ^x/ ^J\ «x» V Our notation may be extended to a function of any number of independent variables. Thus, if u = F{x, y, z), then we have the three partial derivatives du du du^ dF dF dF — -, — , ——J or — — , — ) — • dx dy dz dx dy dz * The constant valaes are eubstitated in the function before differentiating. t Introduced by Jacobi (1801-1851). PARTIAL DIFFERENTIATION 195 Ex. 1. Find the partial derivatives of « = ox^ + 2 bxy + c^. dz * Solution. — =z2ax + 2by, treating ]^ as a constant, dx — = 2 to + 2 c^, treating x as a constant. dy Ex. 2. Find the partial derivatives of u = sin (ox -{-by -{- cz). Solution, — = a cos (ax + by + cz)^ ti-eating y and z as constants, dx — = b cos (ox + 6y + cz), treating x and z as constants, a-, — = c COS lax + &v + <72)) treating y and x as constants. dz Again turning to the function we have by (A) defined — as the limit of the ratio of the increment dx of the function (y being constant) to the increment of a:, as the incre- dz ment of x approaches the limit zero. Similarly (B) has defined — • It is evident, however, that if we look upon these partial deriva- tives from the point of view of § 106, p. 148, then dz dx may be considered as the ratio of the time rates of change of z and X when y is constant, and o^ dy as the ratio of the time rates of change of z and y when x is constant. 135. Partial derivatives interpreted geometrically. Let the equa- tion of the surface shown in the figure (next page) be z =f(x, y). Pass a plane EFGH through the point P (where a; = a and y = J) on the surface parallel to the XOZ plane. Since the equation of this plane is «/ = 6 the equation of the section JPK cut out of the surface is z==f{x, 6), 196 DIFFERENTIAL CALCULUS if we consider EF as the axis of Z and EH as the axis of X. lu this plane — means the same aa dz ^ -=-. and we hare dx — = tan MTP dx = slope of lection JK at P. Similarly, if we pass the plane BCD through P parallel to the YOZ plane, its equation is and for the section DPI, ~ means the same as --• Hence di/ dy ~ = tan MT'P = slope of section 2)1 at P. Ex. 1. Qifen the ellipaoid h --- + — = 1 ; fln<i the slope of the section of the ellipaoid made (a) by the pliuie ]/ = 1 at, the point where z = 4 and z ii poritive; (b) by the plane z = 2 at the point where y = S and z is positive. Solution. Conddering y as constant, 2i 2^ Si _ ^^ _ x_ 24 ~0 ex~ ' °'' tx~~ iz' When z is constant, ?i^ + ^ ^ = 0, or, ^ = _ J^. 12 Q di/ By 2z (a) Wheni/=landi = 4,i = -J^. ..^^-.-J^. Am. (b) Whena! = 2andi* = 3,j: = — . .•. — = ~-Vi. Am. -ft iy 1 ^■Biy+Cy'+Dx + Ey+F. PAETIAL DIFFERENTIATION 197 n / o . V -» . ov- . du 2anzu du _^ 2 6^Jiyu 4. u = arc sin - • -4n«. — = cu X c'y y VyS — X* 5. u = x*. -4?w. — = yzff-^ ; ax au — = xJ'logx. oy 6. u = axV« + tey'«* + cy« + dxz». ilrw. — = 3 ax'^^z + fry*** + d2* ; dx — = 2ax*yz + 3tey«2:* + 6q^; ay a^ 7. u = x^ — 2 xy* + 3 x^y* ; show that x \- y — = 6tt. dx (iy 8. u = — ^— ; show that x h y — = w. X + y ax ay 9. u = (y - z) (2 — x) (x — y) ; show that — + — H = 0. as dy dz 10. u = log (e* + c"') ; show that h — = 1. dx dy 11. u = ; show that 1 = (x + y — l)u. e* + CI' dx dy du du 12. u = xi'y* ; show that x H y — = (x + y + log u) u. dx dy 13. u = log(x« + y» + «»-3xyz); show that — + ^'^ + — = ^ dx dy dz x-\-y-\-z 14. tt = c*8in y + c'' sin x ; show that /?? y + ( — y= «'* + c't' + 2 e'+vsin (x + y). 15. tt = log (tan X + tan y + tan z) ; show that sin 2 X 1- sin 2 y 1- sin 2 z — = 2. dx dy dz 16. Let y be the altitude of a right circular cone and x the radius of its base. Show (a) that if the base remains constant, the volume changes ^irx^ times as fast as the altitude ; (b) that if the altitude remains constant, the volume changes f rxy times as fast as the radius of the base. 198 DIFFERENTIAL CALCULUS 17. A point moves on the elliptic paraboloid z = — f- — and also in a plane par- 9 4 allel to the XOZ plane. When x = 3 f t. and }» increasing at the rate of 9 It. per second, find (a) the time rate of change of z ; (b) the magnitude of the velocity of the point ; (c) the direction of its motion. Ana. (a) r, = 6 f t per sec. ; (b) r = 3 Vis ft. per sec. ; (c) r = arc tan f , the angle made with the XOT plane. 18. If, on the surface of Ex. 17, the point moves in a plane parallel to the plane YOZ, find, when y — 2 and increases at the rate of 5 ft. per sec, (a) the time rate of change of z ; (b) the magnitude of the velocity of the point ; (c) the direction of its motion. ^^ ^^^ ^ ^^ ^^ ^^ ; (b) 6 V2 ft. per sec. ; (c) r = — ) the angle made with the plane XOT, 4. 136. Total derivatives. We have already, on page 57, considered the differentiation of a function of one function of a single inde- pendent variable. Thus, if y =f{v) and v = <^(a;), it was shown that dy ^dy dv dx dv dx We shall next consider a function of two variables, both of which depend on a single independent variable. Consider the function u=/(x, y), where x and y are functions of a third variable t. Let t take on the increment Af, and let A:r, Ay, At^ be the corre- sponding increments of x^ y, u respectively. Then the quantity Aw =f{x -h Aa:, y 4. Ay) -f(x, y) is called the total increment of u. Adding and subtracting /(a:, y 4- Ay) in the second member, {A) Lu = [fix 4- At, y -h Ay) -/(:r, y 4. Ay)] 4- [f{x, y 4- Ay) -/(x, y)] . Applying the Theorem of Mean Value, (44), p. 168, to each of the two differences on the right-hand side of (^), we get, for the first difference, (5) /(a? 4- A2:, y + Ay) -/(a:, y 4- Ay) =/,7a; 4- 5i • A2;, y 4- Ay) Aar. ras X, Aas Aor, and since x varies while .v + Ay remainsl « Lconstant, ve get the partial derirative with respect to x.\ PARTIAL DIFFERENTIATION 199 For the second difference we get ( 0) /{x, y + Ay) -/(x, y) = //(x, y + ^, • Ay) Ay. [as y, Aa aa Ay, and eince y yarles while x remaine oon-l Btant, we get the partial derivative with respect to y.J Substituting (B) and ((7) in (A) gives (p) Au = f^(x 4- ^j . Aa;, y 4- Ay) Lx + //(z, y -f- ^2 • Ay) Ay, where 0^ and 5, are positive proper fractions. Dividing (JP) by A^ (E) ^ =/;(ar + ^, . Aa:, y + Ay) ^ +/;(x, y + ^, . Ay) ^ . Now let A^ approach zero as a limit, then r Since Aj; and Ay converge to xero with A<, we get "1 a/IL*o/-'<* + ^ • ^. y + Ay) -/-'(ar, y), and l*/^**o/»'('» ^ + *« ' Ay)-/,'(x. y), a/IL*o/-'(* + ^ • Aor, y + Ay) -/.'(ar, y), and ^*™J* L. /vX^i y) And /y'(x, y) being asaumed continuouB Replacing /(ic, y) by u in (i^), we get the total derivative ^ ' <l« " aa; de "*" c)y dt ' In the same way, if ^, . •^ w =/(a;, y, 2), and 2;, y, 2 are all functions of ^, we get /CA^ — — — — 4- — ^ -A- — — ^ ^ dt" dx dt'^ dy dt'^ dz dt* and so on for any number of variables.* In (49) we may suppose t = x; then y is a function of x^ and u is really a function of the one variable 2:, giving /«\ dUf _ du du dy dx "doc dy doD * In the same way from (50) we have dx" dx dy dx dz dx* * This Is really only a special case of a general theorem which may be stated as follows : If u is a function of the independent variables x, y, 2, • • •, each of these in turn being a funo* tfon of the independent variables r, «, f, • . ., then (with certain assumptions as to continuity) c^ii du dx du dy du dz dr dx dr dy dr dz dr du du and similar expressioitf hold for , — , etc. 200 DIFFERENTIAL CALCULUS s The student should observe that — and -r- have quite different dx ax dti meanings. The partial derivative — is formed on the supposition ox that the particular variable x aloiie varies^ while du^ limit /Aw\ dx Ax=0\AxJ' where Au is the total increment of u caused by changes in all the variables, these increments being due to the change A2; in the inde- pendent variable. In contradistinction to partial derivatives, — , — are called total derivatives with respect to t and x respectively.* Ex. 1. Given m = sin -» x = c*, y = <* ; find — . y dt ^ . .. du- I X du X X dx , dy .. SolxUum. — = - cos - » — = cos - ; -— = c*, -=^ = 2f. dx y y dy y^ y dt dt du 6^ 6^ Substituting in (49), — = (t — 2) — cos — • Ans. Ex. 2. Given « = ««(y - *), y = asinx, r = cosx; find *?. dx du . du du dy dz Solution. — = ae^ (y - 2), — = e«*, — = — c*^ ; -=^ = a cos x, — = — sin x. dx dy dz dx dx Substituting in (62), du -— = a€f'{y — z) + a€f^coaX'\- e^Binx = e**{a^ + l)sinx. Ans, dx Note. In examples like the above, u could, by substitution, be found explicitly in terms of the independent variable and then differentiated directly, but generally this process would be longer and in many cases could not be used at all. 137. Total differentials. Multiplying (49) and (50) through by dt, we get, (§ 104, p. 144), (58) ^^ "= ^ '^ + ay ^^' (54) au^ — dx + -dy + — dzi o It should be observed that x- has a perfectly definite value for any point (x, y)t vhile — ex dx depends not only on the point (x, y) but also on the particular direction chosen to reach that point. Hence ^^ V- is called a point function ; while dx — is not called a point function unless It is agreed to approach the dx point from some particular direction. PARTIAL DIFFERENTIATION 201 and so on.* Equations (53) and (54) define the quantity du^ which is called a total differential of u or 3, complete differential^ and -r-"^? :r^!/j tt"^ dx dy ^ dz are called partial differentials. These partial differentials are sometimes denoted by d^u^ d^Uy d^u, so that (54) is also written du = d^u 4- d^u 4- rf,w. y Ex. 1. Given u = arc tan - : find du, X* 'Solution, Substituting in (53), tu ^ y bu cy X dx {B» + y« du _^xdy — ydx Ana. x2 + ya Ex. 2. The base and altitude of a rectangle are 6 and 4 inches respectively. At a certain instant they are increasing continuously at the rate of 2 inches and 1 inch per second respectively. At what rate is the area of the rectangle increasing at that instant ? Solution. Let x = base, y = altitude : then u = xy = area, — = y, — = x. dx dy Substituting in (40), , A\ du dx , dy <^> ^=''^+^*- But X = 6 in., y = 4 in., — = 2 in. per sec, - - = 1 in. per sec. dt dt ' du ,'. — = (8 + 5) sq. in. per sec. = 18 sq. in. per sec. Ans, KoTB. Considering du as an infinitesimal increment of area due to the infinites- imal increments dx and dy^ du is evidently the sum of two thin strips added on to the two sides. For, in du = ydx + xdy (multiplying (A) ■^Ji^y^^^y.^^:^. ^ w ^ ' ydx = area of vertical strip, and ^ xdy = area of horizontal strip. But the total increment Au due to the increments dx and dy ^ ^ Att = ydx + xdy + dxdy, Henoe the small rectangle in the upper right-hand comer (— dxdy) 5a is evidently the diif erence between Au and du. This figure illus- trates the fact that the total Increment and the total differential of a function of several vari* ables are not In general equal (see p. 141). * A geometric interpretation of this result will be given on p. 274. r 202 DIFFERENTIAL CALCULUS 138. Differentiation of implicit functions. The equation (A) fix, y) = defines either 2; or y as an implicit function of the other.* It rep- resents any equation containing x and y when all its terms have been transposed to the first member. Let {B) u^f{x,y); du But from (A), f{x, y) =0. .-. w = and — = ; that is, ^ ' dx dy dx Solving for -^» we get dji^ du dy dy a formula for differentiating implicit functions. This formula in the form {C) is equivalent to the process employed in § 75, pp, 83, 84, for differentiating implicit functions, and all the examples on p. 85 may be solved using formula (55). Since n (D) /(^y)=o for all admissible values of x and y, we may say that (55) gives the relative time rates of change ofx and y which keep f(x, y)from chang- ing at all. Geometrically this means that the point {x, y) must move on the curve whose equation is (i>), and (55) determines the direction of its motion at any instant. Since we may write (55) in the form * We assume thai a small change in the yalue of x causes only a small change in the yalue of y t It is assumed that -- and -- exist. ex dy PARTIAL DIFFERENTIATION 203 Ex. 1. 6iyenxV + 8in]^ = 0; find-r^. dx SolvUon, Let/(x, y) = x^ + any, ^ = 2ajy*, ^ = 4xV + co8y. . .froin(66a),^ = - , , ^^ dx 'ay ^"dx 4xV + co8y ^n«. Ex. 2. If X increases at the rate of 2 inches per second as it passes through the value X = 3 inches, at what rate roust y change when y = 1 inch, in order that the function 2xy> — 8xV shall remain constant ? Solution. Let /(x, y) = 2 xy> - 8 x^ ; then ^ = 2y« - 6xy, ^z^^xy-- Ssfi. dx dy Suh8titutingm(66a), dy _ 2y«~6xy dx 4xy — 8x^ »or, dy dif_ Zy^-exy dx d< 4xy - 3x« By (A), p. 164 (56) ft/f dii But X = 8, y = 1, -- = 2. .*. ^ = — 2,^ ft. per second. Ans, dt dt Let F be the point (2;, y, 2) on the surface given by the equation and let PC and AP be sections made by planes through P parallel to the YOZ and XOZ planes respectively. Along the curve -4P, y is constant, therefore from {JE)^ z is an implicit function of x alone, and we have from (55 a) to"" aip' giving the slope at P of the curve -4P, § 183, p. 193. — is used instead of -:;- in the first member since z was originally dx dx from {II) an implicit function of x and y, but (56) is deduced on the hypothesis that y remains constant. Similarly the slope at P of the curve PC is dF (57) ^ = -iy. ^ *' by dF dz 204 DIFFERENTIAL CALCULUS EXAMPLES Find the total derivatiyes, using (49), (50), or (61) in the following six examples. du 1. u = z^ + y^ + zy,z = &inz,y = e*. Ans, -'=3c»*+c»(8inx+cosx)+8in2z. ax du c*(l +x) 2. tt = arc tan (xy). y = €f, Ans. — - = — -^^ — — — • 3. tt = log(a2-p2), p = (,8in^. Ans. — = -2tan^. 4. u = iJ* + cy, « = log », y = 6^. An», ■— - = h tw*. as 9 du « 5. u = arc sin (r — «), r = 3 f, » = 4 1". -An«. — = dt Vl-<i e«f(y — 2) . ^ <*« ^, • 6. u = — ^^^ ^t y = asinx, z = cosx. Ans, --- = e«^smx. a* + 1 dx Using (53) or (54), find the total differentials in the next eight examples. 7. tt = 6y^+ cx« + yy* + «c. Ans. du = (by^-^2cx+e)dx-\'(2byx+Sgy^dy. 8. u = log XV. Ans. dti = - dx + log zdy. X sin X 9. us5«^*. iliM. du = j^* log y COS xdxH dy. 10. u = x^K Ans. du = tt(!^dx + J^^dyV a-^-t A A 2(«tt-«d«) 11. u = . Ans. du = — ^^ — ^ . « - t (« - 0* 12. u = sin (i)g). Ans, du = cos (pg) [gdp + pdq], 13. u = xi^. Ans, du = xff-^ {yzdx + zx log xdy + xy log xdz). 14. u = tan* ^ tana ^tan«f. Ans. du = 4uC-^ + -^ + -^V \sin20 sin2« sin2^/ 15. Assuming the characterlstle equation of a perfect gas to be vp = i?t, where v = volume, p = pressure, t = absolute temperature, and R a constant, what is the relation between the differentials do, dp, dt ? Ans. vdp + pdv = Rdt. 16. Using the result in the last example as applied to air, suppose that in a given case we have found by actual experiment that t = 300® C, p = 2000 lbs. per sq. ft, c = 14.4 cubic feet. Find the change in p, assuming it to be uniform, when t changes to SOP C, and to 14.5 cubic feet. 22 = 90. Ans. - 7.22 lbs. per sq. ft. PARTIAL DIFFERENTIATION 206 In the remaining examples find -^ i using formula (66 a). ox ^ *^ ' ^ ' dx y 2 (x« + ys) + o2 18. e»-e» + xy = 0. -4im. ~ = ^""^ « ^ *> . / V « A ^ ^y y [cos (xy) — c«y — 2 x] 19. sm(xy)-c*i'-x«y = 0. -4)w. -^ = iLL — ^^J'^ i. •dx x[x + e'»'-co8(xy)] 20. sin X sin y + cosx cos y — y = 0. 139. Successive partial derivatives. then, in general, du J du — and — ex dy are functions of both x and y, and may be differentiated again with respect to either independent variable, giving successive par- tial derivatives. Regarding x alone as varying, we denote the results by a^ a^ as^ ru dx"' d2^' dx*' '"' aaf' or, when y alone varies, ^ ^ d^ dTu a/' a/' ay*' '"' ay' the notation being similar to that employed for functions of a single variable. If we differentiate u with respect to a;, regarding y as constant, and then this result with respect to y, regarding x as constant, we obtain — f — b which we denote by • dt/\dxj ^ dydx Similarly, if we differentiate twice with respect to x and then once with respect to y, the result is denoted by the symbol av dydJi?' 206 DIFFERENTIAL CALCULUS 140. Order of differentiation immaterial. Consider the function f(x^ y). Changing x into x + Aa; and keeping y constant, we get from the Theorem of Mean Value, (44), p. 168, (A) f{x + l!ix,y)^f{x,y)^£iX'fJ{x + e^^x,y). 0<^<1 [ass x» Aass Ax, and since x Taries while y remaina oon-l Btant, ^e get the partial derivatiTe with respect to x.\ If we now change y to y + Ay and keep x and Aa; constant, the total increment of the left-hand member of {A) is {B) [f{x + A;r, y + Ay) -f{x, y + Ay)] - [f{x + Aa:, y) -f{x, y)]. The total increment of the right-hand member of {A) found by the Theorem of Mean Value, (44), p. 168, is {O) AxfJ{x + ^Ax, y ^ Ay) - AxfJ {x -^ 0' Ax, y). < ^^ < 1 = AyAxfJ'(x + d,^Ax,y + d^.Ay). < ^, < 1 [ae y, Aas Ajft and since y varies while x and Ax remainl constant, we get the partial deriratiTo with respect to y. J Since the increments {B) and (C) must be equal, (2>) [f{x + Aa:, y + Ay) -f(x, y + Ay)] - [f(x + Ax, y) ^f(x, y)] = AyAxf^^x + e^.Ax, y + ^,. Ay). In the same manner, if we take the increments in the reverse order, {E) [f{x + Aa:, y + Ay) -/(a: + Aa:, y)] ~ [f(x, y + Ay) -f{x, y)] = AxAyf^\x + ^, . Aa;, y + ^, . Ay), 0^ and 0^ also lying between zero and unity. The left-hand members of (2>) and {E) being identical, we have {F) /^> + ^i-Aa:,y + ^,.Ay)=/^"(a: + ^..Aa:,y + ^,.Ay). Taking the limit of both sides as Aa; and Ay approach zero as limits, we have * since these functions are assumed continuous. Placing {(jt) may be written (58) ^'^ ^ dydx dxdy * Assuming the continuity of the first partial derivatlTee and the existence and continuity of ^."and/,.". PARTIAL DIFFERENTIATION 207 That is, the operations of differewtiating with respect to x and with respect to y are commutative. This may be easily extended to higher derivatives. For instance, since (58) is true, da^dy "" dz \dxdy ) " dxdydx dxdy \ dx) dydx \ dx) "" dydj? ' Similarly for functions of three or more variables. Ex. 1. Given u = x»y — Sx^y* ; verify dydx dxdy SolutioTi. — = 8*«y-6xy«, — = 3x2_i8zy2, dx dydx — = x» - 9xV» -^ = 3x« - 18xy«; hence verified. dy dxdy 1. u = cos(x + 2^); 2. u = ?^; 3. tt = ylog(l + 3cy); 4. tt = arc tan - ; 6. u = Bin(^^); 6. u = 6c«y«« + 3c»x»z«+ 2c«x*y - xyz; show that ^J' =12(e'y + ci'z + ex) dx'^cydz 7. tt = e*i" ; show that = (1 + Sxyz + xV**)"- axayaz 8. u = — ^; showthatx— - + y-— - = 2 — . x + y 5x* dxay ax 9. tt = (x« + y*)*; 8howthat8x---- + 3y---+ — = 0. ^ ' axey dy^ dy * V i ^u - '- ' 10. M = y«z%* + z^V + xVc*; show that ^— — — - = e« + e* + A cx^cy-'cz'* 1 c*u dhi dhi 11. u = (x« + y» + «r*;«howthat — + — + — = 0. veriiy dydx dxdy verify dhi dydx dhi dxdy verify dhi dydx dhi dxdy verify dhi df^da dhi 'd8cr^'* verify dhi dedip^ a»tt dipf^de T.1tT • a1 linw thfl. , dhi CHAPTER XVIII ENVELOPES • in. Family ot curves. Variable parameter. The equation of a curve generally involves, besides the variables x and y, certain con- stants upon which the size, shape, and position of that particular curve depend. For example, the locus of the equation (A) ix-af + y^^r' is a circle whose center lies on the axis of X at a distance of a from the origin, its size depending on the radius r. Suppose a to take on a series of values, then we shall have a corresponding series of circles differing in their distances from the origin, as shown in the figure. Any system of curves formed in this way is called a family of curves^ and the quantity a, which is constant for any one curve, but changes in passing from one curve to another, is called a vari- able parameter. As will appear later on, problems occur which involve two or more parameters. The above series of circles is said to be 9, family depending an one parameter. To indicate that a enters as a vari- able parameter it is usual to insert it in the functional symbol, *"^= /(a:,y,a)=0. 142. Envelope of a family of curves depending on one parameter. Any two neighboring curves of a family will in general intersect.* If the corresponding values of the parameter are a and a + Ao, the point or points of intersection, if these exist, will in general tend to definite limiting positions (points) as Aa approaches zero. The locus of all such limiting points is called the envelope of the family * An exception to this would be the Bystem of concentric circles we get from {A) when a is constant and r varies, no two of which would intersect. 208 ENVELOPES 209 of curves. Thus, in the last figure, the limiting positions of the points of intersection of the circles are all on the straight lines AB and CD, which form therefore the envelope of the family of cii'cles. Now find the equation of the envelope of a family of curves depending on one parameter. Let (B) /(x, y, a)=0 and (0) f(x,ff,a + Aa)=0 be two neighboring curves of the same family ^ '(b) intersecting at a point (2/, y') ; and let us find the limiting position of this point of intersection as Aa approaches the limit zero. We can find the equation of a third curve through (a/, y') by applying the Theorem of Mean Value, (44), p. 168, to (B) and ((7), regarding a as the variable and x and y as constants. For we have (D) f{x,y,a-\-Aa)-f{x,y,a) = AafJ(x,t/,a-{-0.Aa). 0<^<1 Since P' lies on both of the curves (B) and ((7), the left-hand members of their equations vanish for x = xf and y = y'. Hence the left-hand member of (2>) must vanish for the same values, and consequently the right-hand member also. Therefore (J?) fJ(x,y,a + 0.Aa)^O is the equation of a third curve passing through the intersection of (B) and ((7). If then (B) and (O) intersect in a point which approaches a fixed point as a limit as Aa approaches zero, we get in general (F) /;(2^y,a)=0 as the equation of a curve which passes through the limit of the intersection of (B) and (C), In general (F) is distinct from (B) and therefore has a definite intersection with it. Since the coordinates of the points on the envelope satisfy both (F) and (jB), its equation is found by eliminating a between these equations. The equation of the envelope is therefore a new rela- tion between x and y that is independent of a.* * By definition we should solve (B) and (C) Bimnltaneously for their point of intersection and then pass to the limit. In practice, however, it is found to he more convenient first to pass to the limit and then solve for x and y, just as we do here. It is not self-evident hy any means that these two processes give the same results in all cases, hut it is a fact that the results are iden- tical in all the applications made in this hook. 210 DIFFERENTIAL CALCULUS On account of the*process of elimination that is involved, no detailed method of procedure can be given for finding the envel- ope that will apply in all cases. In a large number of problems, however, the student may be guided by the following General directions for finding the envelope. First step. . Differentiate with respect to the variable parameter ^ con- siderinff all other quantities involved in the given equation as constants. Second step. Solve the resvltfor the variable parameter. Third step. Svhstitute this value of the variable parameter in the given equation. This gives the equation of the envelope. Ex. 1. Find the envelope of the straight line y = mx + — » where the slope m is the variable parameter. Solviion, First step. o = x — m- 2 Second step. m = ± Third step, y = ± \/| « db -^1 P = ± 2 V^, and squaring, y^ = 4|>x, a parabola, is the equation of the envelope. The family of straight lines formed by varying the slope m is shown in the figure, each line being tangent to the envelope, for we know from Analytic Geometry that i^ = mx + — is the tangent to the parab- ola y' = 4px expressed in terms of its own slope m. 143. The envelope touches each curve of the family at the limiting points on that curve. Q-eometrical proof Let A^ B^ C hQ three neighboring curves of the family, A and B ii^tersecting at P, and B and C at Q. Draw TT' through P and Q. Now let A and C approach coin- cidence with B^ that is, let A^ B^ C become consecutive curves * of the family. Then TT^ becomes a tangent to B^ having two *The limiting position of any point of intersection (as P in figure, p. 211) is sometimes called the point of intersection of two e(m8ee%U%ve curves of the family. Similarly the line which TT* approaches as P approaches Q, i.e. the tangent to iB at Q, is said to pass through two c<m»ee%Uive points of the curve. Of course there is no one curve that is consecutive to another nor any one point that is consecutive to another in the ordinary sense of the word, but geometrical considera- tions have suggested the above phraseolo^v nnd It is understood to be merely a brief way of indicating the actual condition of affairs as stated in the definitions. ENVELOPES 211 consecutive points P and Q in common with it. But then P and Q will also become consecutive points of the envelope by definition ; hence TT' will at the same time become a tangent to the envelope. Therefore B and the envelope have a common tangent ; similarly for every curve of the family. Thus in the example of the last section we noticed that the parabola had the family of straight lines as tangents. Analytical proof. Consider the family of curves represented by (A) f(x, y, «) = 0. dy The 8loi>e at any point on (il) is the value of -^ from dx (B) ?^ + ^J? = 0; (61), p. 190 dx dy dx where, in differentiating, a must be kept constant. From the previous section we know that the envelope of the family of curves is found by eliminating a between {A) and (C) |-/(x, y, a) = 0. da If we suppose (C) solved for a in terms of z and y and the result substituted in (A), it is evident that equation (A) would then be the equation of the envelope. Hence the slope of the envelope may be found by taking the total derivative of (^), (52), p. 199, regarding a as a certain function of x and y determined by (C). This gives dx dy dx da dx Suppose now that the coordinates of the point (x, y) satisfy both (A) and (C) ; that point is therefore on the curve (A) and also on the envelope ; and, by (C), the last term in (D) vanishes, reducing (D) to the same form as {B). Hence at the point (x, y) the slope is the same for the curve {A) and the envelope, so that a limiting point of intersection on any member of the family is a point of contact of this curve with the envelope.* 144. Parametric equations of the envelope of a family depending on one parameter. Instead of finding the equation of the envelope in rectangular form by the method of § 142, p. 210, it is sometimes more convenient to get the equations of the envelope in para- metric form by solving f(x, y, a) = and —f{x, y, a) = for X and y in terms of a. Thus : • In the special caae when ^ = ^ = or -• =0 for all points of oar locus this reasonlr ex cy cy 212 DIFFERENTIAL CALCULUS Ex. 1. Find the envelope of the familyiof straight lines x cos a + y sin a = p, a being the variable parameter. Solution, (A) z cos a + y sin a = p. Differentiating (A) with respect to a, (B) — X sin a + y cos a = 0. Multiplying {A) by cos a and {E) by sin a and sub- tracting, we get , = j,co8.. Similarly, eliminating x between {A) and (B), we ^ 2^ = p sin a. The parametric equations of the envelope are there- fore (C) Iz=p cos a, y = psino; a being the parameter. Squaring equations (C) and adding, we get x« + y« = p«, the rectangular equation of the envelope, which is a circle. Ex. 2. Find the envelope of a line of constant length a, whose extremities move along two fixed rectangular axes. Solution. Let AB = a in length, and let {A) X cos a + ysino— p = be its equation. Now as AB moves always touching the two axes, both a and p will vary. But p may be found in terms of a. For, ^O = ^ B cos a = a cos a, and p = ^ sin a = a sin a cos a. Substituting in (A\ (B) X cos a + y sin a — a sin a cos a = 0, where a is the variable parameter. Differentiating (B) with respect to a, (C) — X sin o -f y cos o + a sin* o — a cos* o = 0. Solving (B) and (C) for x and y in terms of a, we get IX = a sin' a, y = a cos' a, the parametric equations of the envelope, a hypocycloid. The corresponding rectangular equation is found from equations (D) by eliminat ing a as follows : x' = a' sin* a. yi = a* cos* a. Adding, x' + y' = a*, the rectangular equation of the hypocycloid. ENVELOPES 218 146. The evolute of a giveo curve considered as the eovelope of its normals. Since the normals to a carve are all tangent to the evolute, §129, p. 186, it is evident that tU evolute of a carve may also be defined as the envelope of iti normals ; that is, as the locua of the ultimate intersections of neighboring normals. It is also interesting to notice that if we find the parametric equations of the envelope by the method of the previoua section, we get the coordinates x and y of the center of curvatnre ; bo that we have here a aecond method for finding the cod'rdtnates of the center of curvature. If we then eliminate the variable parameter, we have a relation between x and y which ia the rectangular equation of the evolute (envelope of the normals). Solution. The equation of the normal at an; point (z', ^ is '-'■ — Ti^"^ from (2), p. 90. Aa we are considering the normals all along the curre, both z' and y' will vary. Eliminating x' by means of v^ = 4 px', we get the equation of the normal to be Considering j/' as the yariable parameter, we wish to find the envelope of Uiis family of normals. DlBerenti sting (A) with respect to v', _1 = ^_^ Spi 2p and solving for x, Subetitating this value of x in {A) and solving for v, y^-yi. " 4p' (B) sud (C) are then the coordinates of the center of curvature of the parabola. Taken together, (S) and (C) are the parametric equations of the evolute in terms of the paramet«T y". Eliminating y" between (if) and (C) gives 27py* = 4(i-2p)", the rectangular equation of the evolute of the parabola. This is the same result we obtained in Ei. 1, p. 183, by the first method. (C) 214 DIFFERENTIAL CALCULUS 146. Two parameters connected by one equation of condition. Many problems occur where it is convenient to use two parameters connected by an equation of condition. For instance, the example given in the last section involves the two parameters J and y' which are connected hy the equation of the curve. In this case we eliminated x\ leaving only the one parameter y'. However, when the elimination is difficult to perform, both the given equation and the equation of condition between the two parameters may be differentiated with respect to one of the param- eters, regarding either parameter as a function of the other. By studying the solution of the following problem the process will be made clear. s coincide und SobiXvi-n. (A) i- + ^ = 1 la the equRtion of the elllpM where a and b are the variable panuneters connected by the equation (B) iroft = k, tab being the area of an ellipse whose semiaxes are a and h. Differentiating (A.) and (B), regarding a and b Eis vari- ables and X and y ae constants, we have, naing diSereiitjala, —J- + ^ = 0. from {A), and Ma + adb = 0, from (B). Transposing one tenn in each to tbe second member and dividing, we get Therefore, from {A\, —-- and — = - , giving a = ± I V^ and h = ±y^. Substituilng these values in (B), we get the envelope a pair of conjugate rectangular hyperlwlas (see flgora). ENVELOPES 215 1. Find the envelope of the family of straight lines y = 2 mx + m*, m being the Tariable parameter. Ana. x = — 2 m«, y = — 3 m* ; or, 16 y« + 27 z* = 0.» 2. Find the envelope of the family of parabolas ^ = a(x — a), a being the variable parameter. An8. x = 2 a, y = ± a ; or, y = i J x. 3. Find the envelope of the family of circles x^ + (y — /S)^ = r^, /3 being the variable parameter. Ana, x=:±r. 4. Find the equation of the curve having as tangents the family of straight lines y =:mx± Va*wM-^i the slope m being the variable parameter. Ana. The ellipse Wx^ + aV ^ o?t^' 6. Find the envelope of the family of circles whose diameters are double ordi- nates of the parabola y^ = 4px. Ana. The parabola y^ = 4p (p + x). 6. Find the envelope of the family of circles whose diameters are donble ordi- nates of the ellipse V^^ + a^r/^ = a^fta. ^ „^ „. x« y^ ^ Ana. The ellipse — — -- + ^ = 1. 7. A circle moves with its center on the parabola y^ = 4 ox, and its circumference passes through the vertex of the parabola. Find the equation of the locus of the points of ultimate intersection of the circles. Ana. The cissoid y^ (x + 2 a) + x* = 0. 8. Find the curve whose tangents are yz=lx± VaP -|_ ^ + c, the slope I being supposed to vary. Ana. i (ay^ + 6xy + cx^) = 4ac -h^. 9. Find the evolute of the ellipse ¥x'* + a^> = a^b^, taking the equation of normal in the form &y = ax tan - (a«-62)8in 0, the eccentric angle </> being the parameter. -47W. X = — — — COS* 0, y = — -■ — sin* 0; or, (ox)' + (&y)' = (o^— ft^jl. a 10. Find the evolute of the hypocycloid x' + y' = a*, the equation of whose normal is y cos t - x sin t = a cos 2 r, T being the parameter. Ana. (x + y)' + (x - y)' = 2 a'. 11. Find the envelope of the circles which pass through the origin and have their centers on the hyperbola x^ — y^ = ^a. Ana. The lemniscate (x^ + y^)^ = a^ (x^ - y«). 12. Find the envelope of a line such that the sum of its intercepts on the axes equals c. Ana. The parabola x* + y* = c*. * When two answers are given, the first Is in parametric form and the second In rectangnhir form. 216 DIFFERENTIAL CALCULUS Finl the CBvdope of tke fualj of of iiB 14. Find the CBTeiope of becwmn the cqoal to L 15. ftojedilei are fired froMm the Sam hTpocjdoid x^ + y* = A of the A9f^ Aeq gun vishan e»begi in the of an poBible anr 1 .»» trajcctoriea. the and oidi that the is eonstant and adea are (x :^ jr)* = I*. the gun is kefit always is the enielope of the »=*' «r,« CHAPTER XIX SERIES 147. Introduction. A series is a succession of separate numbers which is formed according to some rule or law. Each niunber is called a term of the series. Thus, is a series whose law of formation is that each term after the first is found by multiplying the preceding term by 2 ; hence we may write down as many more terms of the series as we please, and any particular term of the series may be found by substituting the number of that term in the series for n in the expression 2*"*, which is called the general or nth term of the series. In the following six series : (a) Discover by insi)ection the law of formation ; (b) write down several terms more in each; (c) find the nth or genercU term. Series nth term 1. 1,3,9,27,.... s^-i. 2. — a, + a*, — a», +«*,•• •• (- a)". 3. 1, 4, 9, 16, .... n2. . X* X« X* *• ^' 2' 3^ 4^-- X" n B. 4, -2, +1, -i, .... .4(-i).-«. « 3y 5ya 7^3 2 6 10 n»+l *^ Write down the first four terms of each series whose nth or general term is given below. nth term Series 7. fA£\ X, 4x2, 9x», 16x*. X" X x^ x» X* 8 l+Vn ^^^^^ .^^^^^ I + V2 I+V3 1+v^ XT 218 DIFFERENTIAL CALCULUS rUhterm iSeriea Q n-f2 3 4 6 6 2' 9* 28' 65 10. ^. 2» 12 3 4 2' 4' r 16- ^^ (l0g0)»5B» loga-x log^tt'X* log^a-x* log* a* a:* 1 • 2 ' 6 • 24 • |2n-l 148. Infinite series. Consider the series of n terms (A) 1 ^ ^ ^ ^ • W 1. 2' 4' 8 ■■■' 2^*' and let S, denote the sum of the series. Then (5) s, = l + l + l + l + ... + ±,. Evidently S^ is a function of w, for when n = 1, ^1 = 1 =1, when 71 = 2, S^ = l + \ =1J, when M = 3, ^3 = 1 + 14-1 =:1|, when n = 4, ^^ = 1+1 + 1 + 1 =:1J, when n = n^ ^. = 1H-x + t + 7^H h :r— r = 2 — -r— r-* Mark off points on a straight line whose distances from a fixed , point correspond to these different sums. It is seen that the point corresponding to any sum bisects the distance between the preceding point and 2. Hence it appears geometrically that when n increases without limit limit /S. = 2. * Fonnd by 6, p. 1, for the sum of a geometric series. SERIES 219 We also see that this is so from arithmetical considerations, for limit^ Umit/2_ 1 \ 2. I since when n inoreases without limit, ^»proaohes xero as a limit.] We have so far discussed only a particular series (A) when the number of terms increases without limit. Let us now consider the general problem, using the series (C) Mj, u„ u„ u^y ..., whose terms may be either positive or negative. Denoting by S^ the sum of the first n terms, we have /Sf, = Wi + t^j, 4- w, + . . . + w., and ^S; is a function of n. If we now let the number of terms (= n) increase without limit, one of two things may happen : either Case I. 8^ approaches a limit, say it, indicated by Case II. S^ approaches no limit. In either case {O) is called an infinite series. In Case I the infinite series is said to be convergent and to converge to the value ti^ ot to have the value u^ or to have the sum u. The infinite geometric series discussed at the beginning of this section is an example of a convergent series, and it converges to the value 2. In fact, the simplest example of a convergent series is the infinite geometric series a, ar, ar^^ ar^j ar^^ . • ., where r is numerically less than unity. The sum of the first n terms of this series is, by 6, p. 1, „ a(l — r") a af ^' — Trr=i~r-rrr- If we now suppose n to increase without limit, the first fraction * Such a result is sometimes^ for the sake of brevity, called the gum of the series ; hut the student must not forget that 2 is iu>t the sum but the limit of the tumj as the number of terms increases without limit. 220 DIFFERENTIAL CALCULUS on the rightrhand side remains unchanged, while the second approaches zero as a limit. Hence limit « ^ n = ao * 1 — r a perfectly definite number in any given case. In Case II the infinite series is said to be rumconvergent,* Series under this head may be divided into two classes. First class. Divergent series, in which the sum of n terms increases indefinitely in numerical value as n increases without limit; for example, the series from which we get /S.=l+2 + 84---- + w. As n increases without limit, S^ increases without limit and therefore the series is divergent Second class. Oscillating series, of which ^,=1-1 + 1-1+. .. + (-1)-^ is an example. Here S^ is zero or unity according as n is even or odd, and although S^ does not become infinite as n increases without limit, it does not tend to a. limit, but oscillates. It is evident that if all the terms of a series have the same sign the series cannot oscillate. Since the sum of a converging series is a perfectly definite num- ber, while such a thing as the sum of a nonconvergent series does not exist, it follows at once that it is absolutely essential in any given problem involving infinite series to determine whether or not the series is convergent. This is often a problem of great difficulty, and we shall consider only the simplest cases. 149. Existence of a limit. When a series is given we cannot in general, as in the case of a geometric series, actually find the num- ber which is the limit of S^. But although we may not know how to compute the numerical value of that limit, it is of prime importance to know that a limit does exist, for otherwise the series may be nonconvergent. When examining a series to determine whether or not it is convergent, the following theorems, which we state without proofs, are found to be of fundamental importance.! * Some writers nse divergent as equivalent to tionamvergent, t See Osgood's Introduction to Infinite Seriett pp. 4, 14, 04. SERIES 221 Theorem I. If S^ is a variable that always increases as n increases^ hut always remains less than some definite fia^d number A^ then as n increases without limitf S^ will approach a definite limit which is not greater than A. Theorem II. If S^ is a variable that always decreases as n increases^ but always remains greater than some definite fixed number B, then as n increases without limits S^ will approach a definite limit which is not less than B, Theorem III. The necessary and sufficient condition that S^ shall approach some definite fixed number as a limit as n increases without limit is that ,. .. limit .^ _S)=0 for all values of the integer p. 150. Fundamental test for convergence. Summing up first n and then n -f y terms of a series, we have {A) ^« = Wi + w, -f Mj + . . . 4- u^. (B) /S,+^ = tti-fw,-f Ws + '-' + Wn + w.+i + '-' + ^.+p- Subtracting (-4) from (J9), {(^) ^n^p - 'S, = tt,^.i 4- u^^2 + • • • + w„+p. From Theorem III we know that the necessary and sufficient condition that the series shall be convergent is that for every value of p. But this is the same as the left-hand member of ((7); therefore from the right-hand member the condition may also be written Since (B) is true for every value of p, then letting je? = 1, a necessary condition for convergence is that or, what amounts to the same thing. 222 DIFFERENTIAL CALCULUS Hence, if the general (or nth) term of a series does not approach zero as n approaches infinity, we know at once that the series is nonconvergent and we need proceed no further. However, (H) is not a sufficient condition, that is, even if the wth term does approach zero we cannot state positively that the series is convergent ; for, consider the harmonic series .111 1 ' 2 3 4 'n Here j''"'* («,) = »■»" (-\ = 0, that is, condition (E) is fulfilled. Yet we may show that the har- monic series is not convergent by the following comparison: {F) l+J + [J + J] + [J+J + ^+|] + [i + ... + ^] + .... («) i + i+[t + i] + [i + i + i + i] + [^ + - + ^]+--. We notice that every term of {O) is equal to or less than the corresponding term of (-F), so that the sum of any number of the first terms of {F) will be greater than the sum of the corresponding terms of (6r). But since the sum of the terms grouped in each bracket ui.{0) equals ^, the sum of (6r) may be made as large as we please by taking terms enough. The sum ((?) increases indefinitely as the number of terms increases without limit ; hence ((7), and therefore also (^), is divergent. We shall now proceed to deduce special tests which as a rule are easier to apply than the above theorems. 151. Comparison test for convergence. In many cases, an example of which was given in the last section, it is easy to determine whether or not a given series is convergent by comparing it term by term with another series whose character is known. Let {A) Uj4-Wj4-w,4. ... he a aeries of positive terms which it is desired to test for convergence. If a series of positive terms already knotun to be convergent^ namely^ can be found whose terms are never less than the corresponding terms SERIES 228 in the series (A) to be tested^ then {A) is a convergent series and its sum does not exceed that of (B). Proof, Let «« = t^i 4- Wj 4- w, h 1- w^, and ^n = «i + «a + «« 4- • • -f «, ; and suppose that ,. . « — «« '^M = -^• fl = oo " Then, since 'Sn<^ ^^^ *ii ~ '^h'» it follows that s^ < A. Hence, by Theorem I, p. 221, s^ approaches a limit ; therefore the series (-4.) is convergent and the limit of its sum is not greater than A. Ex. 1. Test the series (C) i+i + l + l + 14..... ^ ' 22 3» 4* 6* Solution. Each term after the first is less than the corresponding term of the geometric series 1111 (D) 1 + - + -+- + - + •••, which is known to be convergent (p. 218) ; hence (C7) is also convergent. Following a line of reasoning similar to that applied to (A) and (-B), it is evident that, if (JE) w^ ^. w^ 4. Wj 4. . . . is a series of positive terms to be tested which are never less than the corresponding terms of the series of positive terms, namely , (F) 5^4.6,4.63 + ..., known to be divergent, then (U) is a divergent series. Ex. 2. Test the series ^ . ^ V^ V3 V4 Solution, This series is divergent since its terms are greater than the corre- sponding terms of the harmonic series , 1 1 1 1 + - + - + » .284 ' which is known (p. 222) to be divergent. Ex. 3. Test the following series for different values of p. (G) i4JL + i- + i-4.. . 2p Zp ^P h* a 1 224 DIFFERENTIAL CALCULUS SobUion, Grouping the terms, we have, when p > 1, 2p 8i» 2p 2p^2p 2''-i' 4p S** 61' 7p 4p 4p 4^ 4p 4p \2p-i/ UP Sp Sp Sp SP bP Sp Sp Sp 8p \2i»-i/ and BO on. Construct the series When p > 1, series (H) is a geometric series with the common ratio less than unity, aitd is therefore convergent. But the sum of {G) is less than the sum of {H), as shown by the above inequalities ; therefore (€f) is also convergent. When p = 1, series {G) becomes the harmonic series which we saw was divergent, and neither of the above tests applies. When p < 1, the terms of series {G) vnll, after the first, be greater than the corresponding terms of the harmonic seiies ; hence (G) is divergent. 152. Cauchy's ratio test for convergence. Let (A) u^ + u^^u^^.>. be a series of positive terms to be tested. Divide any general term by the one that immediately precedes it, i.e. form the test ratio -^^ As n increases without limit, let ^^ -2±i = p. W= QO w ' I. When p<l. By the definition of a limit (§ 29, p. 19) we can choose n so large, say w = w, that when n^m the ratio u w+i u shall differ from p by as little as we please, and therefore be less than a proper fraction r. Hence and so on. Therefore, after the term w„, each term of the series (A) is less than the corresponding term of the geometrical series But since r < 1, the series (5), and therefore also the series (-4), is convergent.* * When examining a series for convergence we are at liberty to disregard any finite number of terms ; the rejection of such terms would aif ect the value but not the exUtence of the limit. SERIES 225 II. When p>l {or p= c»). Following tho same line of reason- ing as in I, the series (A) may be shown to be divergent. III. When p = 1, the series may be either convergent or diver- gent; that is, there is no test. For, consider the series ((7) on p. 223, namely, 2" 8' 4f tif {n + ly The test ratio is ^^> = (-^\ = (l ^-Y; and limit /m.+A_ limit A _1 V ,.., ^, . Hence p = 1 no matter what value p may have. But on p. 224 we showed that when jt? > 1, the sei ies converges, and when JE? < 1, the series diverges. Thus it appears that p can equal unity both for convergent and for divergent series, and the ratio test for convergence fails. There are other tests to apply in cases like this, but the scope of our book does not admit of their consideration. Our results may then be stated in compact form as follows: Ghiven the series of positive terms t^i 4- Wg 4- Wj -f . . . -f w„ -f w^+i + • • •; find the limit ^^^^ (^^^ = P- I. When p < 1,* the series is convergent. II. When p > 1, the series is divergent. III. When /> = 1, there is no test. Ex. 1. Test the following series for convergence : c = 1 + i + ,4 + ,4 + ri + • • • + -^ + .- + •••. 1 [2 [3 [4 \n-\ \n * It is not enongh that um+\/un becomes and remains less tlian unity for all yalues of n, but this test requires tbat tbe limit of u»+i/u«i sball be less tban unity. For instance, in the case of the harmonic series this ratio is always less than unity and yet the series diverges as we haye seen. The limit, however, is not less than unity but equals unity. 226 DIFFERENTIAL CALCULUS 8olvti4m. The nth term is ; therefore n — 1 Umit (^^!L±l\= limit / ^ \ ^ limit AlL:zl \ _ limit /l\-or=p) n = oo\ t4„ / n = «l 1 I n = »\ [n / n = co\n/ ^ '' \[nEi/ and by I, p. 225, the series is convergent. \l_ \2_ \S_ Ex.2. Test the series t:: + r^ -^ tzi + " ' - 10 102 io» \J1 Solution. The nth term is here — ; therefore 10" limit /l??i±i\ ^ limit AlL±l ^ 10»\ _ n^a / n + l \ _. ^ n = oo\ Un / n = ao\iOi»+i [n / n = oo\ IQ / ' and by n, p. 226, the series is divergent. Ex. 3. Test the series (O — + — + — +•••. ^ ' 123. 46. 6 Solution, Here the nth term is ; therefore (2n-l)2n limit /«»-n\^ limit [ (2 n ~ 1) 2 n -j ^^ n = oD\ M, / n = <» L(2n + l)(2n4-2)J This gives no test (III, p. 225). But if we comi>are series (C) with (G), p. 223, making p = 2, namely, (D) 1 + — + — H h • • •, we see that (C) must be convergent since its terms are less than the corresponding terms of (2>), which was proven convergent. 153. Alternating series. This is the name given to a series whose terms are alternately positive and negative. Such series occur frequently in practice and are of considerable importance. Jf Wj — w, 4- 1^, — w^ H is an alternating series whose terms never increase in numerical value^ and if ^"^'l u^ = 0, then the series is convergent. Proof. The sum of 2 n (an even number) terms may be written in the two forms (^) 'S,, = (i^i-t^,) + (t4,-w,) + K-We)4-----fK.-i-0»or, SERIES 227 Since each difference is positive (if it is not zero, and the assump- tion ^*^*^ ^« = excludes equality of the terms of the series), series (A) shows that S^^ is positive and increases with w, while series (B) shows that S^^ is always less than u^; therefore by Theo- rem I, p. 221, 5,, must approach a limit less than u^ when n increases, and the series is convergent. Ex. 1. Test the alternating series 1 1 1 . ^ 2 8 4 Solution, Since each term is leas In numerical value than the preceding one, and limit (^)^ limit /I \o. n = oo^ ' n = oo\n/ the series is convergent. 154. Absolute convergence. A series is said to be absolutely* or unconditionally convergent when the series formed from it by mak- ing all its terms positive is convergent. Other convergent series are said to be not absolutely convergent or conditionally convergent. To this latter class belong some convergent alternating series. For example, the series 2» 3« 4* 6* is absolutely convergent since the series ((7), p. 228, namely, ^2* 3« 4* 6* is convergent. The series 2^3 4^6 is conditionally convergent since the harmonic series ^1111. ^2^3^4^6^ is divergent (p. 222). A series with terms of different signs is convergent if the series deduced from it by making all the signs positive is convergent. The proofs of this and the following theorem are omitted. *Tbe tensB of the new series are the Dumerical (absolute) yaluee of the tenns of the glyeo series. 228 DIFFERENTIAL CALCULUS Without placing any restriction on the signs of the terms of the series, the tests given on p. 225 may be stated in the following more general form : Ghiven the series Wi + W« + W, 4- «^4 + • • • -f Wn + W« + 1 + • • • ; calculate the limit ( -=^^ ) = />, limit /!!5iL±i\ = L When\p II. When\p < 1, the series is absolutely convergent. > 1, the series is divergent. III. When |p| = 1, there is no test. 155. Power series. A series of ascending integral powers of a variable, say x^ of the form {A) a^ -f a^x -f- a^ -f a^ H — , where the coefficients a^, a^, a,, . • • are independent of :r, is called a power series in rr. Such series are of prime importance in the further study of the Calculus. In special cases a power series in x may converge for all values of rr, but in general it will converge for some values of x and be divergent for other values of x. We shall examine {A) only for the case when the coefficients are such that limit / a»+A n^co\ a. ) A where Z is a definite number. In {A) lunit '- ^ ^~=^ '- -"-^^ n unit /^,^i\ limit /5i±l^\= li™i^ /^n^A ^r^ia: Referring to tests I, II, III, we have in this case p = Lx, and hence the series (A) is I. Absolutely convergent when \Lz\<ly or \x\< lies between 1 L and -f 1 L 1_ L i.e. when x II. Divergent when \Lx\>\^ or \x\> than — L or greater than + 1 L L i.e. when x is less SERIES 229 1 L ; i.e. when z = dt 1 L III. No test when \Lx\=:ly or \x\ = Note. When Z = 0, it is evident from I (p. 228) that the power series is absolutely convergent for all finite values of 3c Ex. 1. Test the series SolutUm. The series formed by the coefficients is X h 2« 3« 4« 22 ^ 32 42 ^ Here limit /«»±i\ ^ limit r '^^ 1:= limit r/j l_Vl=-l/' = i;\ L\ \-l = 1. By I the series is absolutely convergent when x lies between — 1 and + 1* By II the series is divergent when x \a less than — 1 or greater than + 1. By III there is no test- when x = ±l. But in either case (B) is convergent from the first theorem under § 164, p. 227, since (D), p. 226, was proved convergent. The series in the above example is said to have [— 1, 1] as the irUervcU of conver- gence. This may be v^ritten — 1 < x < 1, or indicated graphically as follows : -1 Show that the following nine series are convergent. 1. l + i + i^- •••. 12 ^ 22 ^ 32 ^ 2. 1 + 1 + 1 + 1+.. 22228 2*^ 4 1 + Ll? + LA:^ + ' 3 36 3.6.9 o. — I — + — + •••• \^ [k \i 2V2 8v3 4V4 7 1-^ + ^ ^ + ^ 8. 1-1. 1 + 1. 1-1. 1 + . 2 2233 2* 42* 9. ' + ' log 2 log 3 log 4 Show that the following three series are divergent. 10.1 + 1 + 1+.... 12 1 + 1 + 2.1 + 3. 1+4 ll.ii + i?+-l4 + 10 102 108 230 DIFFERENTIAL CALCULUS For what values of the variable are the following Graphical representations of series convergent ? *^**™^ ^' oonvergenoe .• [2 • [3 1» 3a 62 Ans, AU values of a. ® -1 41 13. l + x + x* + «•••. Ana, -l<x<l. Q) I ® -1, +1 14. X -- + --- + .... ^TW. -l<x<l. 2 3 4 16. X + X* + aC» + x^' + • • • . -4ns. — 1< x < 1. m I o -¥ +1 16. x + -^4--^ + - •. Am. --1<x<1. ' ■ i Q V2 V3 ^ 5 +1 X^ x' —00 I ^<o 17. 1 + X + — + — 4----. An8. All values of as. ^ ^ ^ ^ — eo I -f-«o 18. 1 1 !-•••. An8. AU values of $. < I > 12 li li 5 ^« ^ «aT -co I -f«o 19. 0-i- + i--f- + .... ^na. All values of 0. ^ I » 15 15 II *» ^^ sin o sin 3 o sin 5 a -«> | +« ^O. — ■ — — — — -p i rt- C08X C08 2X . C08 3x . , _ I -^ 21. \- — h — 1 Ans. X > 0. d i > Bint. Neither the sine nor ooeine can exceed 1 numerically. 22. l4.xloga-f5^ + ?^ + .... Z^ I i? [2 [3 Ana. All values of x. Ill 1 +00 23. + ;; - + ; 1 + --. Ans. X > 1. 1 ® » 1 + X 1 + X2 1 + X« 0+1 ^. 1 x8 1.3 x6 , 1.36 xi . ^ 24. X H H h h • • • . (D I ® 2 3 2-462.4.67 -1 +i Ana. — 1 < X < 1. • End points that are not included in the interval of oonvergenoe have drcles drawn about them. CHAPTER XX EXPANSION OF FUNCTIONS 156. Introduction. The student is already familiar with some methods of expanding certain functions into series. Thus, by the Binomial Theorem, (A) (a -f a:)* = a* + 4 a"a: 4- 6 aV -f 4 oa?" -f a^, giving a finite power series from which the exact value of (a -f- x)* for any value of x may be calculated. Also by actual division, 1 — X \l — xy we get an equivalent series, all of whose coefficients except that of of are constants, n being a positive integer. Suppose we wish to calculate the value of this function when X = .5, not by substituting directly in 1 but by substituting x = .5 in the equivalent series Assuming n = 8, ((7) gives for a; = .5 (D) . zA— =1.9921875 + .0078125. 1 — a: If we then assume the value of the function to be the sum of the first eight terms of series ((7), the error we make is .0078126. However, in case we need the value of the function correct to two decimal places only, the number 1.99 is as close an approximation to the true value as we care for since the error is less than .01. It is evident that if a greater degree of accuracy is desired, all we need to do is to use more terms of the powe^ series {H) l+a;-fa:*4-a:*H 231 232 • DIFFERENTIAL CALCULUS Since, however, we see at once that 2, Li-J.-." there is no necessity for the above discussion, except for purposes of illustration. As a matter of fact the process of computing the value of a function from an equivalent series into which it has been expanded is of the greatest practical importance, the values of the elementary transcendental functions such as the sine, cosine, logarithm, etc., being computed most simply in this way. So far we have learned how to expand only a few special forms into series; we shall now consider a method of expansion appli- cable to an extensive and important class of functions and called Taylor's Theorem. 157. Taylor^s Theorem* and Taylor^s Series. Replacing J by a: in (j&), p. 169, the extended theorem of the mean takes on the form (59) A^) = Aa) + ^^^^r(a) + i^^/Fr(a) + ^^^^^fnf^a) + • • • Li I* Is where x^ lies between a and a*. (59), which is one of the most far- reaching theorems in the Calculus, is called Taylor^B Theorem. We see that it expresses /(a:) as the sum of a finite series in {x — a). (x — aY The last term in (59), namely, ^— ; — -f^^Mt is sometimes called |n the remainder in Taylor's Theorem after n terms. If this remainder converges towards zero as the number of terms increases without limit, then the right-hand side of (59) becomes an infinite power series called Taylor^ Series,^ In that case we may write (59) in the form (60) f(x) = f(a) + ^^^^r{a) + <^^/rf (a) + ^^^ f^a) + • • ., LL L* 15 and we say that the function has been expanded into a Taylor^ s Series. For all values of x for which the remainder approaches zero as n increases without limit, this series converges and its sum gives the * AIbo known as Taylor*i Formula. t Published by Dr. Brook Taylor (1685-1731) in his Methodut IncremetUorum^ London, 1715. EXPANSION OF FUNCTIONS 233 exact value of /(x), because the difference (= the remainder) between the function and the sum of n terms of the series approaches the limit zero (§ 30, p. 21). On the other hand, if the series converges for values of x for which the remainder does not approach zero as n increases without limit, then the limit of the sum of the series is not equal to the function /(a:). The infinite series (60) represents the function for those values of x and those only for which the remainder approaches zero as the num- ber of terms increases without limit. It is usually easier to determine the interval of convergence of the series than that for which the remainder approaches zero ; hut in simple cases tlie two intervals are identical (see footnote, p. 236). When the values of a function and its successive derivatives are known for some value of the variable, as 2; = a, then (60) is used for finding the value of the function for values of x near a, and (60) is also called the expansion of f{x) in the vicinity of x=za. Ex. 1. Expand log x in powers of (x — 1). Solution. f(x) = log X, /(I) = ; r(x) = -, /'(1) = 1; X X' Substituting in (60), logx = x -1 -i(x - l)« + i(x - 1)» . Ana. Ttiis converges for values of x between and 2 (§ 166, p. 228) and is the expan- sion of log X in the vicinity o/x = 1, the remainder converging to zero. When a function of the sum of two numbers a and x is given, say f{a -f x), it is frequently desirable to expand the function into a power series in one of them, say x. For this purpose we use another form of Taylor's Series, gotten by replacing a; by a -f a: in (60), namely, (61) f(a + x)= f(a) + jf /'(a) + ,^/"(a) + %rHa) + • • •. 11 15 Iff 234 DIFFERENTIAL CALCULUS Ex. 1. Expand sin (a + x) in powers of x. SoiutioTL Here f{a + x) = Bin (a + x) ; hence, placing x = 0, /(a) = sin a, /'(a) = COB a, /"(a) = - sin a, /'"(a) = - COS a, Substituting in (61), X x^ x^ sin (a 4- x) = sin a + - COB a — -- sin a — -- COS a + • • • . J.1W. EXAMPLES* 1. Expand e* in powers of X- 2. Ans. c* = c» + ^(x - 2) + — (x - 2)«+ •• •. I? 2. Expand x* — 2x^ + 5x — 7 in powers of x — 1. Ana. -3 + 4(x-l) + (x-l)«+(x-l)". 3. Expand 3v*~14y + 7in powers of y — 3. Am. -8 + 4(y-3) + 3(y-3)«. 4. Expand 5z3 + 7z + 3in powers of z — 2. Ana. 37+27(«-2) + 6(«-2)«. 6. Expand cos (a + x) in powers of x. as Ana. cos(a + x) = cos a — xsin a — — cos a H — sin a + • • •. |2 [3 6. Expand log (x + h) in powers of x. 2 ^a Ana. log(x + A) = log;i + ^-^ + — + .... 7. Expand tan (x + h) in powers of h. Ana. tan (x + A) = tan x + A sec^ x + A'sec* x tan x + • • • . 8. Expand the following in powers of h. (a) (x + A)'> = x" + nx''-'^ + ''^^"'^^ x-«ft8 + ^^'*"'|^^'^^^^ x"-8A8 + "-> [2 [3 (l,)e«+* = c«(l + A + | + ^ + ...). [2 li 158. Maclaurin's Theorem and Maclaurin's Series. A particular case of Taylor's Theorem is found by placing a = in (59), p. 232, giving (62) fix) = /(O) + ^ /'(O) + ^ fff(0) + ^ /'"(O) + . . . Lt I* lil + if— r /<"-"' (O) + ^ /^"> («!). * In these examples we assume that the functions can be dereloped Into a power series. EXPANSION OF FUNCTIONS 236 where x^ lies between and x. (69) is called Maclaurin's Theorem. The right-hand member is evidently a series in a; in the same sense that (59)9 p. 232, is a series in x — a. Placing a = in (60), p. 232, we get MaclaurirCs Series^* (63) f(x) =/(0) + j^/'(0) + 2!/fr(0) + ^/'"(O) + •.., Li I* 12 a special case of Taylor's Series that is very useful. The state- ments made concerning the remainder and the convergence of Taylor's Series apply with equal force to Maclaurin's Series, the latter being merely a special case of the former. The student should not fail to note the importance of such an expansion as (63). In all practical computations results correct to a certain number of decimal places are sought, and since the process in question replaces a function perhaps difficult to calcu- late by an ordinary polynomial with constant coefficients^ it is veiy useful in simplifying such computations. Of course we must use terms enough to give the desired degree of accuracy. In the case of an alternating series (§ 153, p. 226) the error made by stopping at any term is numerically less than that term, since the sum of the series after that term is numerically less than that term. Ex. 1. Expand cos x into an infinite power series and determine for what values of X it converges. Solution, Differentiating first and then placing x = 0, we get f(x) = cos X, /(O) = 1, /'(x) = ~sinx, /'(0) = 0, r(x)=-cosx, r(0)=-l, /"' (X) = sin X, /'" (0) = 0, /^▼(x) = cosx, /iv{0) = l, /^(x) = -sinx, ' /^(0) = 0, /Ti(x) = - cosx, /^(O) = - 1, etc., etc. Suhstitnting in (63), X* X* a^ (A) C08X = 1--+--- + .... • Named after Colin Maclanrln (1698-1746), being first published in his Treatise o/Fluxiant, Edinbuigh, 1743. The Series is really due to Stirling (1091M770). 236 DIFFERENTIAL CALCULUS Comparing with Ex. 18, p. 230, we see that the series converges for ail values of as. In the same way for sin x. gj8 2^ 2"7 (B) «nx = «-- + --- + .... which converges for all values of x (Ex. 10, p. 230).* Ex. 2. Using the series {B) found in the last example, calculate sin 1 correct to four decimal places. SoltUion, Here x = 1 ; that is, the angle is expressed in circular measure (see second footnote, p. 17). Therefore, substituting x = 1 in (B) of the last example, Summing up the positive and negative terms separately, 1 = 1.00000- . • .- = 0.16667. . • i = 0.00833- • . .- = 0.00019- . • ti L7 1.00833' - 0.16686--' Hence sin 1 = 1.00833 - 0.16686 = 0.84147 - - • , which is correct to four decimal places since the error made must be less than — } i.e. less than .000003. Obviously the value of sinl may be calculated to any desired degree of accuracy by simply including a sufficient number of additional terms. EXAMPLES Verify the following expansions of functions into power series by Maclaurin^s Series and determine for what values of the variable they are convergent. jpfl jg' x^ 1. ^ = 1 + aj + — -f — - + — i . Convergent for all values of x. L± l£ Lz x^ x^ sfi x' 2. co8x = 1 —,-- + ■- — — + — •. Convergent for all values of X. [2 [4 |6 [8 • Since here /('•)(x)=ain (x + ^) and /(»)(xt)«»ain (x^ + — ) , we have, by subetitating in the lai»t term of (62), p. 234, ^^ 2 / \ 2/ remainder = — Bin I XiH )• 0<Xi<ar [n \ 2 / . But sin (xi + — ) can never exceed unity, and from Ex. 17, p. 230, Wmit — =o for all values \ 2 / w=qo|n of X. Hence ijmif ar" . / nn\ ^ limit _8in(x, + ^)=0 n=ao[n \ * 2/ for all ralues of x ; that is, in this case the limit of the remainder i» for all valuea of x for which the series converges. This is also the case for all the functions considered in this book. EXPANSION OF FUNCTIONS 287 3. «. = i + xlog, + ^l^ + ^i^ + 12 [8 4. sin ke = KX h !-■••. 13 16 II ^ 1 /^ V X' X» 3C* X* 6. log (1 + x) = X h • • '^^^ 23 46 « , .- . X* x« X* X* 7. log (1 — x) = — X 8. arc sin X = X H h h • • • . 23 245 x' x^ x^ sfi 9. arc tan x = x 1 — -\ •.-. 3 6 7 9 10. 8in2x = x«-- — + — +•••. L3 [6 11. erfii«fr = i^ 0^r__zL-|- .... 2 8 9" 4^ 8^ 12. c«8in^=^+^+^- ---^ . 3 ^ Conyergent for all values of x. Convergent for all values of x, k being any constant. • Convergent for all values of x, k being any constant. Convergent if — 1 < x < 1. Convergent if ~ 1 < x < 1. Convergent if — 1 < x < 1. Convergent if — 1 < x < 1. Convergent for all values of x. Convergent for all values of 0. Convergent for all values of 0. 13. Show that log x cannot be expanded by Maclaurin^s Theorem. Compute the values of the following functions by substituting directly in the equivalent power series, taking terms enough until the results agree with those given below. 14. 6 = 2.7182... SohUUm. Let x = 1 in series of Ex. 1 ; then c = l + H--H 1 !-—+•••. First term = 1.00000 Second term = 1.00000 Third term = 0.50000 Fourth term = 0.16667.. Fifth term = 0.04167 . . Sixth term = 0.00833.. Seventh term = 0.00139.. Eighth term _ = 0.00019 . . •, etc. Adding, e = 2.71826.. Ans, [DiTiding third term by 3.] [Dividing fourth term by 4.] [IMriding fifth term by 5.] [Dividing sixth term by 6.] [Dividing seventh term by 7.] 15. arc tan (I) = 0.1973 • > . ; use series in Ex. 9. 16. cos 1 = 0.5403 • • . ; use series in Ex. 2. 17. cos 10° = 0.9848 • • • ; use series in Ex. 2. 288 DIFFERENTIAL CALCULUS 18. sin 7 = 0.7071 • • • ; iwe series (B), p. 28a 4 19. sin .6 = 0.4794 • • • ; use series (B), p. 236. 20. c9 = 1 4- 2 + - + .^ + • • • = 7.3891. [2 [3 169. Computation by series. I. 2%e computation of ir hy series. From Ex. 8, p. 287, we have arc sin x^=X'\ h 1- . . •. ^2.8 2.4.5 2.4.6.7 Since this series converges ♦ for values of x between — 1 and + 1, we may let a; = i, giving 6 2^2 3V2y ^2.4 5V2y ^ • or, 7r = 3.1415 ... Evidently we might have used the series of Ex. 9, p. 237, instead. Both of these series converge rather slowly, but there are other series, found by more elaborate methods, by means of which the correct value of tt to a large number of decimal places may be easily calculated. II. The computation of logarithms by series. Series play a very important r81e in making the necessary calcu- lations for the construction of logarithmic tables. From Ex. 6, p. 287, we have (A) log(l + x) = z-- + --- + ^-.... This series converges for z = 1, and we can find log 2 by placing X = 1 in (A), giving log2 = l-^ + J-} + J-J+.... But this series is not well adapted to numerical computation, because it converges so slowly that it would be necessary to take • We asaume that it oonverges to the correct value. EXPANSION OF FUNCTIONS 239 1000 tenns in order to get the value of log 2 correct to three deci- mal places. A rapidly converging series for computing logarithms will now be deduced. By the theory of logarithms, {B) logl±^ = log(l + n:)-log(l~n:). 8, p. 2 1 — X Substituting in (B) the equivalent series for log (1 -fa;) and log(l — a;) found in Exs. 6 and 7 on p. 237, we get* which is convergent when x is numerically less than unity. Let (2>) --il^ = --:, whence x^ "" , 1 — x N M-\-N and we see that x will always be numerically less than unity for all positive values of M and N. Substituting from (2>) into ((7), we get (E) log^ = logJf-logJ\r a series which is convergent for all positive values of M and N ; and it is always possible to choose M and ^ so as to make it con- verge rapidly. Placing Jf = 2 and JV = 1 in {U)^ we get l°g2 = 2[l-Hl.i + l.i, + l.i + ...]=0.69314718.... [since log N^ log 1-0, and ~ « - • 1 L Jo + iV S J Placing Jf = 8 and iV= 2 in (J?), we get logS =log2 + 2 n + i.i + ^.i + .. ."I =1.09861229.... *The student shonid notice that we have treated the series as if they were ordinary stuns, but they are not ; they are limits of sums. To justify this step is beyond the scope of this book. 240 DIFFERENTIAL CALCULUS It is only necessary to compute the logarithms of prime numbers in this way, the logarithms of composite numbers being then found by using theorems 7-10, p. 2. Thus, log 8 = log 2» == 3 log 2 = 2.07944154 • • ., log 6 = log 8 + log 2 = 1.79175947 .... All the above are Naperian or natural logarithms^ i.e. the base is e = 2.7182818. If we wish to find Brigg%^ or common logarithmsy where the base 10 is employed, all we need to do is to change the base by means of the formula Thus, w 9=.igg^=0'693...^ ^'' log. 10 2.802... In the actual computation of a table of logarithms only a few of the tabulated values are calculated from series, all the rest being found by employing theorems in the theory of logarithms and various ingenious devices designed for the purpose of saving work, EXAMPLES Calculate by the methods of this article the following logarithms. 1. log^S =1.6094.... 3. loge 24 = 3.1781.... 2. logelO = 2.3026 ... 4. logio 5 = 0.6990 .-• . 160. Approximate formulas derived from series. In the two pre- ceding sections we evaluated a function from its equivalent power series by substituting the given value of 2; in a certain number of the first terms of that series, the number of terms taken depend- ing on the degree of accuracy required. It is of great practical importance to note that this really means that we are considering the function as approximately equal to an ordinary polynomial with constant coefficients. For example, consider the series {A) Binx = x-| + |-^ + .... This is an alternating series for both positive and negative values of X. Hence the error made if we assume sin a; to be approximately EXPANSION OF FUNCTIONS 241 equal to the sum of the first n terms is numerically less than the (n -f l)th term, § 153, p. 227. For example, assume (B) sin x = x^ and let us find for what values of x this is correct to three places of decimals. To do this, set (0) < .001. This gives x numerically less than V.006 (= .1817); that is, {B) is correct to three decimal places when x lies between -{- 10° .4 and - 10°.4. The error made in neglecting all terms in (A) after the one in i"~' is given by the remainder, (62), p. 234, m E = ^&\x,); \n hence we can find for what values of 2: a polynomial represents the function to any desired degree of accuracy by writing the inequality (E) \E\ < limit of error, and solving for a-, provided we know the maximum value off^*\x^). Thus, if we wish to find for what values of x the formula (F) sin a; = a: — — i; correct to two decimal places (i.e. error <. 01), knowing that |/«^>(a:i)|<l, we have from (2>) and (H), <M; that is,la:|<'V^; or, |a:|<l. 120 a:" Therefore x-r-^ gives the correct value of sin x to two decimal D places if |a^|< 1? i.e. if x lies between + 57° and — 57°. This agrees with the discussion of (A) as an alternating series. Since in a great many practical problems accuracy to two or three decimal places only is required, the usefulness of such approximate formulas as {B) and (F) is apparent. 242 DIFFERENTIAL CALCULUS Muygens^ approximation to the length of a circular arc. Let S denote the length of the required arc, C its chord, c the chord of half the arc, and B the radius of the circle. The circular measure of the whole angle being S £ 1 , . S 2 . S 2 we have sin g-g = ;^' ^^^Ib^H' ^^' C . S c . S = sin--— 1 •r--=sin 2Ii 2R 2B ^B Developing the right-hand members of the last two equations into power series in S by (68), p. 235, we get C 8 S" . S^ . (fl^) 2B 2B 1 28. 2»^^ 12. 5. 2*^ \ f OJ? AT? 10Q0«J?8'' 2R ^B l-2.3.2«^» • 1.2...6.2"^ Multipl3dng {H) by 8 and then subtracting ((?) from {H) in order to eliminate the term containing S\ we have approximately 8g-(7 ^3^ 3 S^ 2B ~'2B 4 12 .62*^' 8c-C „ S' or, — - — = 5 — 3 7680 -K* Hence, for an arc equal in length to the radius, the error in taking is less than j^-q of the whole arc. For an arc of half the length of the radius the proportionate error is one sixteenth less, and so on.* In practice Huygens' approximation is generally used in the form {J) S = 2c + ^{2c-C). • For an angle of 30* the error is less than 1 In 100,000, for 46* leas than 1 in 20,000, and for 00 less than 1 In OOOO EXPANSION OF FUNCTIONS 243 This simple method of finding approximately the length of an arc of a circle is much employed. To find the approximate length of a portion of any continuous curve, divide it into an even number of suitable arcs, regarding the arcs as approximately circular. Ex. 1. Find by Huygens* approximation the length of an arc of 30^ in a circle whose radius is 100,000 ft. Solution. Here c = 2 B sin 7° 30', C = 2 B sin 16® ; but from tables of natural sines we get sin 7° 30^ = .1306268, sin 16° = .2688190. Substituting in {J), a =^52369.71. The true value, assumuig r = 3.1416926, is 62369.88 ; hence the error is but .17 ft., or about 2 inches. EXAMPLES 1. Draw the graphs of the functions x, x , z — — H — respectively, and compare them with the graph of sin x. L L l_ 2. If d is the distance between the middle points of the chord c and the circular arc a, show that the error in taking _8^ . , ,, 82 d* ^ " * 3 s IS less than . 3 a» 161. Taylor's Theorem for functions of two or more variables. The scope of this book will allow only an elementary treatment of the expansion of functions involving more than one variable by Taylor's Theorem. The expressions for the remainder are complicated and will not be written down. Having given the function it is required to expand the function {B) /{x+h.y + k) in powers of h and k. Consider the function (C) f(x + U,y + kt). Evidently {B) is the value of {€) when < = 1. Considering (C) as a function of t, we may write (D) fix + ht,y + kt)= F(t), 244 DIFFERENTIAL CALCULUS which may then be expanded in powers of t by Maclaurin's Theo- rem, (62), p. 234, giving (H) F{t) = F{0) + tF^ (0) + ^ F''{0) + 1 F^''(0) + • • • . Let us now express the successive derivatives of F{t) with respect to ^ in terms of the partial derivatives of F{t) with respect to x and y. Let (F) a = x-\-ht, /3 = f/-\-kt; then by (49), p. 199, But from (F), j,n^.dFdadFdfi ^ ^ da dt dfi dt^ -rr = h and -^szk; dt dt and since F{t) is a function of x and y through a and /8, dF^dFda , dF_dFdfi^ dx dadx dy~ dfi dy' or, since from (-F), — = 1 and — = 1, ^ ' dx dy {I) dF dF J dF dF — = — and — = — • dx da dy d/3 Substituting in (Cr) from (/) and (S), (J) ^' dx dy Replacing F{t) by F'(f) in (J"), we get dx dy ^^^kSl^^k dx dxdy [ dxdy^ dfj (K) ... F"{t) = h'll+2hk?^ + ¥^^ dx' ■ dxdy In the same way the third derivative is dy". ^"'(.) = ..g^3.«*^-^ + 3M'^ + *«0, and so on for higher derivatives. EXPANSION OF FUNCTIONS 246 When t = 0, we have from (2>), (6?), (J'), (JT), (X), ^(0)=/(^» y)» i-e« J^(0 ^ replaced \yjf\x, y), ^ ' dx dy ^•"(0) = A* ^ + 2 AA -^ + A* ^,, i?""(0) = A» ^ + 3 A»A -^ + 3 AA* -^. + A» ?^, ^ ' aa^ aar'ay aj«/ ay and so on. Substituting these results in {E), we get («4) /(x + M, y + ^) =/(aB, ») + «(* ^ + * ^) To get /(a: + A, y + A), replace t by 1 in (M), giving Taylor's Theorem for a function of two independent variables, («5) f{!c + h,v + k)=:f{x,v) + h^ + k^ + ;^U,^^ + 2Hk^ + k'm+..., |2\ cte« aasay Oy*/ which is the required expansion in powers of h and k. Evidently (65) is also adapted to the expansion of f{x -j- A, y -j- A:) in powers of X and y by simply interchanging x with h and y with i. Thus, («5o) f(x + h,y + k)=f(h,k) + x^^ + y^ +^K^+''^^+»'^)+ |2\ a;k« • '^ aMit ' " dk* Similarly for three variables we shall find (66) f{x + h,y + k,z + l)=f(x,y,z) + h^ + k^ + l^( ^I2V c>x"^ c)f/»^ c)«>^ [2V 5x' c)y» dz* dxdy ' d;edx dydz ' and so on for any number of variables. 246 DIFFERENTIAL CALCULUS EXAMPLES 1. Given /(x, y) = Ax^ + Bxy + Cy" ; expand /(x + A, y + *) In powera of h and ib. Solution, — = 2Ax + By, — =JBx + 2Cy; dx ^ dx^ dxdy ay« The third and higher partial deriyatives are all zero. Sabstituting in (65), /(x + A, 1/ + *) = -4x2 + Bxy + Cj/» + (2 -4x + By) A + (BiB + 2 Cy)* + Ah^ + B^iifc + Cifc». An*. 2. Given /(x, y, «) = ^x« + By^ + Cz^ ; expand /(x + i, y + m, « + n) in powerg of {| m, n. 5o/tt<ion. — = 2^x, — = 2By,^ = 2Cz; dx ' dy az ex2 c>y2 az« axay eyaz dzdx The third and higher partial derivatives are all zero. Substituting in (66), /(x + ^ y + rn, z-f n) = -4x« + 5y2+ Cz« + 2 -4xZ + 2 Bym + 2 Czn + -4Z2 + Bma + Cn^, Ana. 3. Given /(x, y) = Vx tan y ; expand/(x + A, y + ^) in powers of h and k, 4. Given /(x, y, z) = -4xa -f By« + Cz^ + Dxy + ^yz + l^zx ; expand /(x + A, y -^ kf z + in powers of h, k^ I. 162. Maxima and minima of functions of two independent vari- ables. The function /(2:, y) is said to be a maximum at 2; = a, y = i when /(a, b) is g^reater than f{x^ y) for all values of x and y in the neighborhood of a and b. Similarly /(a, b) is said to be a minimum at 2; = a, y = i when /(a, 6) is less than f(x^ y) for all values of x and y in the neighborhood of a and &. These definitions may be stated in analytical form as follows : If, for all values of h and k numerically less than some small positive quantity, (A) f{a + A, b-^k) —f{a^ 6) = a negative number^ then /(a, b) is a maximum value of /(a:, y). If (5) f{a -t- A, 6 -f- A:) —f(a^ b)=:a positive number^ then /(a, 6) is a minimum value of /(x, y). EXPANSION OF FUNCTIONS 247 These statements may be interpreted geometrically as follows: a point P on the surface ^. . is a maximum point when it is ^^ higher" than all other points on the surface in its neighborhood, the coordinate plane XOY being assumed horizontal. Similarly P' is a minimum point on the surface when it is «* lower " than all other points on. the surface in its neigh- borhood. It is there- fore evident that all vertical planes through P cut the surface in curves (as AFB or DFE in the figure), each of which has a maximum ordinate z {^MP) at P. In the same manner all vertical planes through P' cut the surface in curves (as BP^C or FP^O)y each of which has a minimum ordinate z {z^NP^) at P'. Also, any contour (as HIJK) cut out of the surface by a horizontal plane in the immediate neighborhood of P must be a small closed curve. Similarly we have the contour LSBT near the minimum point P'. It was shown in §§ 93, 94, pp. 117-121, that a necessary con- dition that a function of one variable should have a maximum or a minimum for a given value of the variable was that its first derivative should be zero for the given value of the variable. Similarly for a function f(x^ y) of two independent variables, a necessary condition that /(a, b) should be a maximum or a minimum (i.e. a turning value) is that for a: = a, y = 6, ^ = 0, ^ = 0. dx dy Proof. Evidently {A) and (P) must hold when A = ; that is, f{a-\.Kh)^f{a,l) is always negative or always positive for all values of h sufficiently small numerically. By §§ 93, 94, a necessary condition for this is {C) 248 DIFFERENTIAL CALCULUS that -r-f(x^ b) shall vanish for a: = a, or, what amounts to the same ax thing, '^rf^'i V) ^^ vanish for a: = or, y = 6. Similarly {A) and ox (B) must hold when A = 0, giving as a second necessary condition that — f{Xy y) shall vanish for a; = a, y = ft. In order to determine sufficient conditions that /(a, ft) shall be a maximum or a minimum it is necessary to proceed to higher derivatives. To derive sufficient conditions for all cases is beyond the scope of this book.* The following discussion, however, will suffice for all the problems given here. Expanding f(a -j- A, ft -h A:) by Taylor's Theorem, (65), p. 245, replacing a; by a and y by ft, we get ox cy -|("S--S-*-|)--. where the partial derivatives are evaluated for a; = a, y = ft, and R denotes the sum of all the terms not written down. All such terms are of a degree higher than the second in A and k. Since — = and r^ = 0, from (C), p. 247, we get, after transpos- dx oy ing/(a, ft), (^) fia + k^h + k) -fia, h) = 1 (A«g -f 2M^ + *»!) + B. If /(a, ft) is a turning value, the expression on the left-hand side of (U) must retain the same sign for all values of A and k suffi- ciently small in numerical value, the negative sign for a maximum value [(J.), p. 246] and the positive sign for a minimum value [(S), p. 246], i.e. /(a, ft) will be a maximum or a minimum accord- ing as the right-hand side of (U) is negative or positive. Now E is of a degree higher than the second in A and k. Hence as A and k diminish in numerical value it seems plausible to conclude that the numerical value of B will eventually become and remain less than • See Courg d 'Analyact Vol. I, by CL Jordan. EXPANSION OF FUNCTIONS 249 the numerical value of the sum of the three terms of the second degree written doum on the right-hand side of (E),* Then the sign of the right-hand side (and therefore also of the left-hand side) will be the same as the sign of the expression OTT. dxdy dy But from Algebra we know that the quadratic expression always has the same sign as A {or B) when AB — C^> 0. Applying this to (F\ A = —,^ B = -^, C = — =^» and we see that ^^•^ ^ ^ ^ dJ^ df dxdy {F)j and therefore also the left-hand member of (H), has the same sign as g(or^) when ay ^ _ /_^Y^ cj^ dt/^ \dxdyj Hence the following rule for finding maximum and minimum values of a function /(;r^ y). First step. Solve the simultaneous equations ^ = 0,^ = 0. dx dy Second step. Calculate for these values of x and y the value of gygy /gyy dJ? df \dxdyj ' Third step. The function vrill have a maximum ?/ A > and ;— f or t^ J < 0; minimum if A>0 and ^ ^or ^)>0; neither a maximum nor a minimum ij A < 0. The question is undecided i/ A = O.f •Peano has shown that this conclusion does not always hold. See the article on *< Maxima and Minima of Functions of Several Variables," by Professor James Pierpont in the BiUletin of the American Mathematical Society ^ Vol. IV. t The discussion of the text merely renders the given rule plausible. The student should observe that the case A = is omitted in the discussion. 250 DIFFERENTIAL CALCULUS The student should notice that this rule does not necessarily give all maximum and minimum values. For a pair of values of X and y determined by the first step may cause A to vanish, and may lead to a maximum or a minimum or neither. Further investigation is therefore necessary for such values. The rule is, however, sufficient for solving many important examples. The question of maxima and minima of functions of three or more independent variables must be left to more advanced treatises. Ex. 1. Examine the fonction Saxy — x* — y^ for maximam and minimum yalues. Solutum, f{z, y) = 8 axy — x* — y*. First step. ^ = 3ay - Sx* = 0, — = 3aa; -3y« = 0. bx dy Solving these two eqaations simultaneously, we get X = 0, X = a, y = 0; y = a. Second step. -4 = -6x, — ^ = 3a, -4=-6y; dz^ dxdy ey« '^ 5x« dy^ \dzdv/ dy^ \dxdy Third step. When x = and y = 0, A = — a^, and there can be neither a maximum nor a minimum at (0, 0).. ay When X = a and y^^a, A = + 27 a^ ; and since -^ = — 6 a, we haye the condi- tions for a maximum yalue of the function fulfilled at (a, a). Substituting x = a, y = a in the giyen function, we get its maximum yalue equal to a*. Ex. 2. Divide a into three parts such that their product shall be a maximum. Solution, Let x = first part, y = second part ; then a — {z-\-y)z:::a^x-'y = third part, and the function to be examined is /(x, y) = xy{a-x-'y). First step. -^=zay - 2xy - y^ - o, — = ox - 2xy - x« = 0. dx By Solving simultaneously, we get as one pair of values x = - « y = -.* 3 8 Secondstep. -^ = - 2y, — ^ = a — 2x — 2y, — = — 2x: axa dxdy dy^ A = 4xy - (a - 2x - 2y)2. ** a?" 0, y » are not considered, since from the nature of the problem we would then have a minimum. EXPANSION OF FUNCTIONS 251 Third step. When a; = - and y = -» A = — : and since — = — --, it is seen 3 8 3 ' 8x« 3 that our product is a maximum when x = - > y = -. Therefore the third part is also - , and the maximum value of the product is — • 3 27 EXAMPLES 1. Find the maximum value oix^ + xy -^ j/* — ax — by. Ans. i (oft — a* — 6"). 3ir 2. Show that sin 2 + sin y 4- cos (x 4- y) is a minimum when x = y = — * and tr ^ a maximum when x = y = — • 3. Show that x&f-^'^y has neither a maximimi nor a minimum. lax A-hu A- c\^ 4. Show that the maximum value of ^ „ * is a« + fts + c^. x3 + ya + 1 6. Find the greatest rectangular parallelopiped that can be inscribed in an ellipsoid. That is, find the maximum value of 8 xyz(= volume) subject to the condition a-a y* «« ^ %aJbc a« 62 c2 3V8 HitU. Let ic >■ xys, and substitute the value of « ftom the equation of the ellipsoid. This gives where « is a function of only tvo variables. 6. Show that the surface of a rectangular purallelopiped of given volume is least when the solid is a cube. 7. Examine x* 4- y* — 2^ 4- xy -> y> f or maximum and minimum values. Ara. Maximum when x = 0, y = ; minimum when x = y = ± i, and when x = — y = i: i V3. 8. Show that when the radius of the base equals the depth, a steel cylin- drical standpipe of a given capacity requires the least amount of material in its construction. 9. Show that the most economical dimensions for a rectangular tank to hold a given volume are a square base and a depth equal to one half the side of the base. 10. The electilc time constant of a cylindrical coil of wire is mxyz ax-\-oy -\- cz where x is the mean radius, y is the difference between the internal and external radii, z is the axial length, and m, a, 6, c are known constants! The volume of the coil is fixyz = g. Find the values of x, y, z which make u a minimum if the volume of the coil is fixed. zlabcg Am. ax=zby^cz=\^ ' CHAPTER XXI ASYMPTOTES. SINGULAR POINTS. CURVE TRACING 163. Rectilinear asymptotes. An asymptote to a curve is the limiting position* of a tangent whose point of contact moves off to an infinite distance from the origin.f Thus, in the hyperbola, the asymptote AB is the limiting position of the tangent FT as the point of contact F moves off to the right to an infinite distance. In the case of algebraic curves the follow- ing definition is useful : an asymptote is the limiting position of a secant as two points of intersection of the secant with a branch of the curve move off in the same direction along that branch to an infinite distance. For example, the asymptote AB is the limiting position of the secant FQ as F and Q move upwards to an infinite distance. 164. Asymptotes found by method of limiting intercepts. The equation of the tangent to a curve at {x^^ y^ is by (1), p. 89, First placing y = and solving for a:, and then placing a;= and solving for y, and denoting the intercepts by x^ and y, respectively, we get dx Xi = a:r^ — y^ - ^ = intercept an OX; y^=y^ — Xj -7^ = intercept on OY. * A line that approaches a fixed straight line as a limitinp; position cannot be wholly at infinity ; hence it follows that an asymptote must pass within a finite distance of the origin. It is evident that a curve whicli has no infinite brancli can have no real asymptote. t Or, less precisely, an asymptote to a curve is sometimes defined as a tangent whoee point of contact is at an infinite distance. 252 ASYMPTOTES 253 Since an asymptote must pass within a finite distance of the origin, one or both of these intercepts must approach finite values as limits when the point of contact {x^, y^ moves off to an infinite distance. If limit (a-,) = a and limit (y,) = 6, then the equation of the asymptote is found by substituting the limiting values a and b in the equation a h If only one of these limits exists, but then we have one intercept and the slope given, so that the equation of the asymptote is y = mx-\-b or a: = — -f a. Ex. 1. Find the asymptotes to the hyperbola — = 1. « , ^. dy K^x 6 1 , limit /dy\ h SoltUion. — = — = ± , and 7»= """^ ( -^ ) = ±-. dx a^y a I ^ ^ = QoVdx/ a Also, a^ = — and y. = ; hence these intercepts are zero when x = y = ao. Therefore the asymptotes pass through the origin (see figure on p. 252) and their eqoations are y — = ± - (x — 0), or, ay = ±bx. Ana. vw This method is frequently too complicated to be of practical use. The most convenient method of determining the asymptotes to algebraic curves is given in the next section. 165. Method for determining asymptotes to algebraic curves. Given the algebraic equation in two variables, (A) fix, y) = 0. If this equation when cleared of fractions and radicals is of degree n, then it may be arranged according to descending powers 254 DIFFERENTIAL CALCULUS m of one of the variables, say y, in the form For a given value of a^ this equation determines in general n values of y. Case I. To determine the asymptotes to the curve (B) which are parallel to the coordinate axis. Let us first investigate for asymp- totes parallel to OF. The equation of any such asymptote is of the form (C) x = k, and it must have two points of intersection with (B) having infinite ordinates. First, Suppose a is not zero in (£), that is, the term in y" is present. Then for any finite value of 2:, (B) gives n values of y, all finite. Hence all such lines as ((7) wiU intersect (B) in points having finite ordinates, and there are no asymptotes parallel to OY, Second, Next suppose a = but ( and c are not zero. Then we know from Algebra that one root (= y) of (B) is infinite for every finite value of x; that is, any arbitrary line (O) intersects (B) at only one point having an infinite ordinate. If now in addition 6a: -f <? = 0, or, then the first two terms in (B) will drop out, and hence two of its roots are infinite. That is, (2)) and (B) intersect in two points having infinite ordinates, and therefore (J9) is the eqtuUion of an asymptote to (B) which is parallel to OY. Third. If a = 6 = <? = 0, there are two values of x that make y in {B) infinite, namely, those satisfying the equation {E) dar»-f ex-f/=0. Solving (E) for x, we get two asymptotes parallel to OYj and so on in general. * For ufle in this section the attention of the student Is called to the following theorem from Algebra : Given an algebraic equation of degree n, ^y" + J5y" -» + Cy" ->+ JDy»- « + •••= 0. When A approaches zero, one root (value of y) approaches «. When A and B approach zero, two roots approach oo. When Af B, and C approach zero, three roots approach oo, etc. ASYMPTOTES 255 In the same way, by arranging /(a:, y) according to descending powers of x^ we may find the asymptotes parallel to OX. Hence the following rule for finding the assrmptotes parallel to the coor- dinate axes : First step. Equate to zero the coefficient of the highest power of x in the equation. This gives all asymptotes parallel to OX, Second step. Etpiate to zero the coefficient of the highest power of y in the eqaatum. This gives all asymptotes parallel to OY. Note. Of course if one or both of these coefficients do not involve x (or y), they cannot be zero, and there will be no corre- sponding asymptote. Ex. 1. Find the asymptotes of the carre aH = y (x -^ a)*. Solution. Arranging the terms according to powers of z, yx« - (2 ay + a^a; + a^ = 0. Equating to zero the coefficient of the highest power of x, we get y = as the. asymptote parallel to OX. In fact the asymptote coincides with the axis of x. Arranging the terms according to powers ^^^' (x-a)«y-a«x =0. Placing the coefficient of y equal to zero, we get X = a twice, showing that ^B is a double asymptote parallel to OY. If this curve is examined for asymptotes oblique to the axes by the method explained below, it will be seen that there are none. Hence y = and x = a are the only asymptotes of the given curve. Case II. To determine asymptotes oblique to the coordinate axes. Given the algebraic equation (F) /(^y)=o. Consider the straight line (Q) y=zmx-\-k. It is required to determine m and k so that the line ( 69^) shall be an asymptote to the curve (F), Since an asymptote is the limiting position of a secant as two points of intersection on the same branch of the curve move off to an infinite distance, if we eliminate y between (F) and ((?), the resulting equation in rr, namely, (H) f{x,mx-^h)^Q, 256 DIFFERENTIAL CALCULUS must have two infinite roots. But this requires that the coeffi- cients of the two highest powers of x shall vanish. Equating these coefficients to zero, we get two equations from which the required values of m and h may be determined. Substituting these values in ((?) gives the equation of an asymptote. Hence the following rule for finding asymptotes oblique to the coordinate axes : First step. Replace y by mx-\-k in the given equation and expand. Second step. Arrange the terms according to descending powers of X. Third step. Equate to zero the coefficients of the two highest powers* of Xy and solve for m and Ic, Fourth step. Substitute these values of m and k in y = mx -f k. This gives the required asymptotes. Ex. 2. Examine ^ = 2 ox^ - x^ for asymptotes. Solution, Since none of the terms involve both x and y, it is evident that there are no susymptotes parallel to the coordinate axes. To find the oblique asymptotes^ eliminate y between the given equation and y = mx -\- k. This gives (?nx + A:)« = 2ax«-z»; and arranging the terms in powers of x, (1 + m8)x' + (3 m«ik - 2 a)x2 + 3 k^mx + *» = 0. Placing the first two coefficients equal to zero, 1 + m* = and 3 rn-^k - 2 a = 0. 2a Solving, we get ?7i = — 1 , A: = Substituting 2a in y = 7HX + fc, we have y=z—x-\ , the equa- tion of asymptote AB. EXAMPLES Examine the first eight curves for asymptotes by the method of § 164, and the remaining ones by the method of § 165. 1. y = e*. Ans. y = 0. 2. y Arts, y = 0. * If the term involving t"~H8 missing, or if tbe value of m obtained by placing the first coefficient equal to zero causes the second coefficient to vanish, then by placing the ooefflcieuts of x'* and x^~2 equal to zero we obtain two equations from which the values of m and k may be found. In this case we shall in general obtain two k*s for each m, that Is, pairs of parallel oblique asymptotes. Similarly, if the term in x**-^ is also missing, each value of m furnishes three parallel oblique asymptotes, and so on. ASYMPTOTES 257 3. y = logx. 1\* 6. y =z tan x. 1 An8, x = 0. n being any odd integer, x 6. y = c* - 1. X = 0, y = 0. 7. y» = 6x« + x». y = x-|-2. 8. Show that the parabola has no a^mptotes. 9. y* = a' — x». y + X = 0. UT 10. The cisBoid y^ = 11. y«a = yax + x». x« 2r-x 12. y«(xa + l) = x«(x»-l). 13. y2(x~2a) = x«-o». 14. xV = a«(x? + y«). 15. y(z*-36x + 26«) = x'-3ax« + a«. 16. y = c4- o« x = 2r. * X = a. y = :iX. x = 2a, y = ± (X + a). x = ±a, » = ± a. x = 6, x = 26, y + 3a = X + 3&. y = c, x = b. y + X + a : = 0. - y = 0. x = 0, y = 0, X + y = 0. (X - 6)» 17. The folium x» + y« - 3axy = 0. 18. The wltchx^y = 4a«(2a-y). 19. xi/<^ -\- x"^ := a\ 20. x» + 2x«y-xy*-2y» + 4y« + 2xy + y = 1. x + 2y = 0, x + y = l, x-y = -l. 166. Asymptotes in polar coordinates. Let /(/>, 0) = be the equation of the curve FQ having the asymptote CD. As the asymptote must pass within a finite distance (as OU) of the origin^ and the point of contact is at an infinite dis- tance, it is evident that the radius vector Oi^ drawn to the point of contact is parallel to the asymptote, and the subtangent OE is perpendicular to it. Or, more precisely, the distance of the asymptote from the origin is the limiting value of the polar subtangent as the point of contact moves off an infinite "distance. 268 DIFFERENTIAL CALCULUS To determine the asymptotes to a polar curve, proceed as follows : First step. Find from the equation of the curve the values of 6 which make p — oo.* These values of give the directions of the asymptotes. Second step. Find the limit of the polar subtangent as 6 approaches each such value^ remembering that p approaches oo at the same time. Third step. If the limiting value of the polar suhtangent is finite^ there is a corresponding asymptote at that distance from the origin and parallel to the radius vector drawn to the point of contact. When this limit is positive the asymptote is to the rights and when negative^ to *he left of the origin^ looking in the direction of the infinite radius vector. EXAMPLES 1. Examine the hyperbolic spiral p = - for asymptotes. V SohUion, When ^ = 0, p = oo. Also — = — — , hence d$ e» d0 a^ 0^ Subtangent = p2— = — . = — <L * *^ dp 0^ a ^= ^~] = " ^' ^^^^^ ^ ^^*®- It happens in this case that the subtangent is the same for all values of e. The curve has therefore an asymptote BC parallel to the initial line OA and at a distance a above it. Examine the following curves for asymptotes. 2. pcos9 = acos2^. Ans. There is an asymptote perpendicular to the initial line at a distance a to the left of the origin. 3. p = a tan B, AnA. There are two asymptotes perpendicular to the initial line and at a dis- tance a from the origin, on either side of it. 4. The lituus p^ = a. Am, The initial line. * If the equation can be written as a polynomial in p, these Talnes of 9 may be found I9 equating to zero the coefficieDt of the highest power of p (see footnote, p. 254). SINGULAR POINTS 259 6. p = a sec 2 ^. Ans. There are four asymptotes at the same distance - from the origin, and inclined 45'' to the initial line. ^ 6. (p ~ a) sin tf = 6. Ana, There is an asymptote parallel to the initial line at the distance b above it. 7. P=a (sec 2 ^ + tan 2 0). An9, There are two asymptotes parallel to 9 = -» at the distance a on each side of the origin. 8. Show that the initial line is an asymptote to two branches of the curve ^sin^ = a3cos2^. 9. Parabola p ?— . An,. There is no asymptote. 1-C08« 167. Singular points. Given a curve whose equation is f{x,t,) = 0. Any point on tlie curve for which ^ = Oand^ = dx dy is called a %ingular point of the curve. All other points are called ordinary points of the curve. Since by {55 a), p. 202, we have dy _ dx dy it is evident that at a singular point the direction of the curve (or tangent) is indeterminate, for the slope takes the form -• In the next section it will be shown how tangents at such points may be found. 168. Determination of the tangent to an algebraic curve at a given point by inspection. If we transform the given equation to a new set of parallel coordinate axes having as origin the point in ques- tion on the curve, we know that the new equation will have no constant term. Hence it may be written in the form {A) f{x, y)=ax + by-{-{ex^-{- dry -f ef) -\-{f^'\-g^y +^^f + «y) + ••• = 0, 260 DIFFERENTIAL CALCULUS The equation of a tangent to the curve at the given point (now the origin) will be (5) y = @)^- By (1), p. 89 Let y^mx (by 54 (c), p. 3) be the equation of a line through the origin and a second point F on the locus of {A). If then F approaches along the curve, we have from {B) (0) limit w = $^. ax Let be an ordinary point. Then, by § 167, a and b do not both vanish since at (0, 0), from (A), dx dy Replace y in (A) by mx^ divide out the factor x^ and let x approach zero as a limit. Then (A) will become* a 4- Am = 0. Hence we have from (-B) and (0) aa; -f 6y = 0, the equation of the tangent. The left-hand member is seen to consist of the terms of the first de^ee in {A). When is not an ordinary point we have a = 6 = 0. Assume that c, (2, e do not all vanish. Then proceeding as before (except that we divide out the factor x'), we find, after letting x approach the limit zero, that {A) becomes c -f dm -f eni? = 0, or, from (0), <^ -<f)-(i)'-»- * After dividing by x an algebraio equation In m remains whose coefficients are f nnotlons of x. If now X approaches zero as a limit, the theorem holds that one root of this equation in m will approach the limit — o-r 6. SINGULAR POINTS 261 Substituting from (£), we see that (H) C2? -f dxy -f e^ = IS the equation of the pair of tangents at the origin. The left- hand member is seen to consist of the terms of the second degree in (A). Such a singular point of the curve is called a double point from the fact that there are two tangents to the curve at that point. Since at (0, 0), from (-4), d3? dxdy df ' it is evident that {D) may be written in the form ^^ dj?^ dxdy\dx)^df\dx) In the same manner, if there is a triple point at the origin, the equation of the three tan- gents being fj? -f g2?y -f hxy"" -{-iy'' = 0. And so on in general. If we wish to investigate the appearance of a curve at a given point, it is of fundamental importance to solve the tangent prob- lem for that point. The above results indicate that this can be done by simple inspection after we have transformed the origin to that point. Hence we have the following rule for finding the tangents at a given point. First step. Transform the origin to the point in question. Second step. Arrange the terms of the resulting equation accord- ing to ascending powers of x and y. Third step. Set the group of terms of lowest degree equal to zero. This gives the equation of the tangents at the point {origin). 262 DIFFERENTIAL CALCULUS Ex. 1. Find the eqaation of the tangent to the ^^ 6x« + 5y2 + 2xy-12aj-12y = at the origin. Solution. Placing the terms of lowest (first) degree equal to zero, we get -12x-12y = 0, or, X + y = 0, which is then the equation of the tangent PT at the origin. Ex. 2. Examine the curre 3 x^ — xy — 2 y^ + £' — 8 y* = f or tangents at the origin. SoliUion. Placing the terms of lowest (second) degree equal to zero, 3x«-xy- 2 2^ = 0, or, (x-y)(8x + 2y) = 0, X — y = being the eqiiation of the tangent AB, and 3 X + 2 y = the equation of the tangent CD. The origin is, then, a double point of the curve. Since the roots of the quadratic equation (F), p. 261, namely, df \dx) dxdy \dx) d3? may be real and unequal, real and equal, or imaginary, there are three cases of double points to be considered, according as ^ ^ \dxdy) d^dy" is positive, zero, or negative (see 3, p. 1). 169. Nodes. ( \dxdy) ax* dy^ In this case there are two real and unequal values of the slope = -T-) found from (jP), so that we have two distinct real tangents to the curve at the singular point in question. This means that the curve passes through the point in two different directions, or, in other words, two branches of the curve cross at this point. Such a singular point we call a real double point of the curve, or a node. Hence the conditions to be satisfied at a node are fee i/)-o ^~o ^-o (J!L.y^^f^f dx^dy^ >0. SINGULAR POINTS 268 Ex. 1. Examine the lemniscate y* = x^ — x* for singular points. Solution. Here /(x, y) = y*--x*4.x* = 0. Also, ^ = -2x + 4x8 = 0, ^ = 2y = 0. dx by The point (0, 0) is a singular point since its coordinates satisfy the above three equations. We have at (0, 0), g = -2. -^=0,^ = 2. ax* bxby ay« \bxdy/ dx^dy^ ' A' and the origin is a double point (node) through which two branches of the curve pass in different directions. By placing the terms of the lowest (second) degree equal to zero we get y2 — xa = 0, or y = X and y = — x, the equations of the two tangents AB and CD at the singular point or node (0, 0). na Cusps. (j?!Ly_^^^ = o. In this case there are two real and equal values of the slope found from (-F), hence there are two coincident tangents. This means that the two branches of the curve which pass through the point are tangent. When the cui-ve recedes from the tangent in both directions from the point of tangency, the singular point is called a point of oscillation ; if it recedes from the point of tangency in one direction only, it is called a ciisp. There are two kinds of cusps. First kind. When the two branches lie on opposite sides of the common tangent Second kind. When the two branches lie on the same side of the common tangent.* The following examples illustrate how we may determine the nature of singular points coming under this head. Ex. 1. Examine a*j/^ = a%c* — sfi for singular points. Solution. Here /(x, y) = aV - a^x* + x« = 0, ^ = -4a^8 + 6x» = 0, ^ = 2a*y = 0, dx dy • Meaning in the neighborhood of the Binguliir point. 264 DIFFERENTIAL CALCULuS and (0, 0) is a singular point since it satisfies the above tliree equations. Also, at (0, 0) we have ay 7} e*^ ax« dxdy ^■^=0.^ = 20.. ey2 and since the curve is symmetrical with respect to OF, the origin is a point of osculation. Placing the terms of lowest (second) degree equal to zero, we get y^ = 0, showing that the two common tangents coincide with OX. Ex. 2. Examine ^ = x* for singular points. Solution. Here /(x, y) = y^ - x» = 0, ^ = -3x» = 0, ^ = 2y = 0, dx dy showing that (0, 0) is a singular point Also, at (0, 0) we have \bxdyJ cx^dy^ ex* ' (XOy = o.|?C = 2. ../^:^y-^.^ = o. ays This is not a point of osculation, however, for if we solve the given equation for y, we get y=iVx8, which shows that the curve extends to the right only of OT^ for negative values of X make y imaginary. The origin is therefore a cusp, and since the branches lie on opposite sides of the common tangent it is a cusp of the first kind. Placing the terms of lowest (second) degree equal to zero, we get y^ = 0, showing that the two common tangents coincide with OX. Ex. 3. Examine (y — x^)^ = x' for singular points. SolMiUm. Proceeding as in the last example, we find a cusp at (0, 0), the common tangents to the two branches coinciding with OX. Solving for y, y = X* ± x'. If we let X take on any value between and 1, y takes on two different positive values, showing that in the vicinity of the origin both branches lie above the common tangent. Hence the singular point (0, 0) is a cusp of the second kind. 171. Conjugate or isolated points, f ^^ V- ^ ^ < o. In this case the values of the slope found from (B) are imagi- nary. Hence there are no real tangents; the singular point is the real intersection of imaginary branches of the curve, and the SINGULAR POINTS 265 coordinates of no other real point in the immediate vicinity satisfy the equation of the curve. Such an isolated point is called a conjugate point, Ex. 1. Examine the carve |/^ = x* — £> for singular points. Solution. Here (0, 0) is found to be a singular point of the curve /hi at which ~ = ± V— 1. Hence the origin is a conjugate point. Solv- dx ing the equation for y, O T y = ± aj Vx-1. This shows clearly that the origin is an isolated point of the curve, for no values of x between and 1 give real values of y. 172. Transcendental singularities. A curve whose equation in- volves transcendental functions is called a transcendental curve. Such a curve may have an end pointy at which it terminates abruptly, caused by a discontinuity in the function ; or a salient point at which two branches of the curve terminate without having a common tangent, caused by a discontinuity in the derivative. Ex. 1. Show that ^ = x log x has an end point at the origin. Solution. X cannot be negative since negative num- bers have no logarithms; hence the curve extends only to the right of OT, When x = 0, y = 0. There being only one value of y for each positive value of x, the curve consists of a single branch terminating at the origin, which is therefore an end point. Ex. 2. Show that y = j- has a salient point at the origin. 1 + c* Solution. Here — = dx 7 + 1 1 + c* x(l + ci)« If X is positive and approaches zero as a limit, we have ultimately y = and -^ = 0. dx If X is negative and approaches zero as a limit, we get ultimately dy 2/ = and -- = 1. dx Hence at the origin two branches meet, one having OX as its tangent and the other, ABj making an angle of 46*^ with OX. 266 DIFFERENTIAL CALCULUS EXAMPLES 1. Show that ^ = 2 xs + X* has a node at the origin, the slopes of the tangents being db V5. 2. Show that the origin is a node of ^(a^ + x*) = x»(a» - x>), and that the tangents bisect the angles between the axes. 3. Prove that (a, 0) is a node of y^ = x (x — a)', and that the slopes of the tangents are db^^* 4. Prove that a^^ — 2 aJbz^ —ofi^O has a point of osculation at the origin. 6. Show that the curve ^ = x^ + x* has a point of osculation at the origin. x' 6. Show that the cissoid y^ — has a cusp of the first kind at the origin. 2a — X 7. Show that ^ = 2 ox^ — x' has a cusp of the first kind at the origin. 8. In the curve (y — x^Y = ^^ show that the origin is a cusp of the first or second kind according as n is < or > 4. 9. Prove that the curve x* — 2 ax*y — oxy* + cfiy'^ = has a cusp of the second kind at the origin. 10. Show that the origin is a conjugate point on the curve y^ (x^ — a^ = x*. 11. Show that the curve y> = x (a + x)^ has a conjugate point at (— a, 0). 12. Show that the origin is a conjugate point on the curve ay^ — x* + bx^ = when a and h have the same sign, and a node when they have opposite signs. 13. Show that the curve x* + 2 ox^ — ay* = has a triple point at the origin, and that the slopes of the tangents are 0, + V^, and — v^. 14. Show that the points of intersection of the curve (-) +(-) =1 with the axes are cusps of the first kind. ^ 15. Show that no curve of the second or third degree in x and y can have a cusp of the second kind. _i 16. Show that y = e ' has an end point at the origin. 17. Show that y = x arc tan - has a salient point at the origin, the slopes of the T * tangents being ± - • 173. Curve tracing. The elementary method of tracing (or plotting) a curve whose equation is given in rectangular coordi- nates, and one with which the student is already familiar, is to solve its equation for y (or a:), assume arbitrary values of x (or y), calculate the corresponding values of y (or x), plot the respective CURVE TRACING 267 points, and draw a smooth curve through them, the result being an approximation to the required curve. This process is laborious at best, and in case the equation of the curve is of a degree higher than the second, the solved form of such an equation may be unsuit- able for the purpose of computation, or else it may fail altogether, since it is not always possible to solve the equation for y or x. The general form of a curve is usually all that is desired, and very often we care to examine the curve in the neighborhood of a certain point only. To attain this object it is as a rule only neces- sary to determine some of the important points, lines, and proper- ties of the curve as enumerated below. No rules for tracing a curve can be given that will apply in all cases, but the student will find it to his advantage to use the fol- lowing general directions as a guide and to study carefully the examples that are worked out in detail. 174. General directions for tracing a curve whose equation is given in rectangular coordinates. 1. Examine the curve for symmetry. (a) If the equation is unchanged when y is replaced by — y, the curve is symmetrical with respect to OX, (b) If the equation is unchanged when x is replaced by — a:, the curve is symmetrical with respect to F. (c) If the equation is unchanged when x is replaced by — a:, and y by — y, the curve is symmetrical with respect to the origin which is also the center of the curve. 2. Examine the curve for important points, (d) If the equation is satisfied by a; = 0, y = 0, the curve passes through the origin. (e) Placing x = and solving for y gives the intercepts on OY. Placing y = and solving for x gives the intercepts on OX, (f) Find -^; this gives the direction of the curve at any point and serves to locate maximum and minimum points (§ 94, p. 120). (g) Find -7^; this gives the direction of curvature at aP" point and serves to find the points of inflection (§ 98, p. 137 (h) Examine the curve for singular points (p. 259). 268 DIFFERENTIAL CALCULUS 3. (i) Uxamine the curve for OBymptotes (§§ 164, 165, p. 252). Determine on which side of each asymptote the corresponding infinite branch lies. 4. (j) Locate additional points on the curve. If possible, com- pute a sufficient number of points on the curve by the elementary method (§ 173, p. 266) to give a fair idea of the locus, and sketch the curve through the points. Ex, 1. Trace the curve y^ = sb*. Solution, Let us examine the curve in the above order. (a) The curve is symmetrical with respect to OX. (b) The curve is not symmetrical with respect to OT, (c) The curve is not symmetrical with respect to the origin. (d) It passes through the origin. (e) Its intercepts on the axes are both zero. y > (f) -p = — - ; showing that above OX the curve always has a positive slope, and below OX a negative slope. It has no maximum or minimum points. 3 hence ^bove OX the curve Is concave up- wards and below OX concave downwards. There are no points of inflection. (h) The curve has a cusp of the first kind at the origin, the common tangent coinciding with OX, (i) There are no asymptotes. (i) y = ± V^; hence the curve does not extend to the left of OT, since nega- tive values of z make y imaginary. When x = oo, y = db oo, showing that there are two infinite branches, one on each side of OX. Plotting a number of points and sketching in the curve, we get the semicubical parabola shown in the figure. Ex. 2. Trace the curve y' = 2 ox^ — x". Solution. This curve is found to be not symmetrical vnth respect to either axis or the origin, but it passes through the origin and in addition has the intercept 2 a 4a. safe uii \y.^L.. dy 4 ox -3x2 V dx 3y2 ' hence when x = 4a 3 the curve has the maximum ordinate } a ^i. d^j 8aa ^^ 9xi(2a-x)*' hence x = 2 a gives a point of inflection on OX, to the left of which the curve is concave down- wards and to the right concave, upwards. The curve has a cusp of the first kin'd at the origin, CURVE TRACING 269 the common tangent coinciding with OT. The only asymptote is (^£ in figure) which lies to the right of the infinite branch in the second quadrant and to the left of the infinite branch in the fourth quadrant. From y = V2 ax^ — x* we plot addi- tional points and draw the curve shown in the figure. 175. Tracing of curves given by equations in polar coordinates. The rudimentary method is to solve the equation for p when pos- sible, assume values for 0, calculate the corresponding values of p, plot the points thus found, and draw a curve through them.* This work may be facilitated by examining the curve for asymp- totes and by noting the values of which make p a maximum or a minimum. Ex. 1. Trace the curve p = 10 sin 2 $, Solution. Tabulating p. 4), we have Bin2e P 0^ 16<» .60 6.0 80° .87 8.7 460 1.00 10.0 GOP .87 8.7 lb"" .60 6.0 90*» 106° - .60 - 6.0 120° - .87 - 8.7 136° -1.00 -10.0 160° - .87 - 8.7 166° - .60 - 6.0 180° 196° .60 6.0 210° .87 8.7 226° 1.00 10.0 240° .87 8.7 266° .60 6.0 270° 286° - .60 - 6.0 800° - .87 - 8.7 316° -1.00 -10.0 330° - .87 - 8.7 346° - .60 - 6.0 360° the corresponding value of $ and p for every 16* (see table, p= 10sin2^. p is a maximum when sin 2 ^ is a maximum, and this occurs when sin 2^ = 1, or = 46°, 226°, etc. This maximum value of p is then 10. p is a minimum when sin 2 tf is a minimum, i.e. when sin 20 = — 1 or 0= 136°, 816°, etc. Hence the minimum value of p is - 10. When = 0, 180°, etc., p = 0. If we in addition remember that sin 2 is a periodic continuous function of 0, it is not neces- sary to tabulate many values of and p. The curve consists of four loops, as shown in the figure, and it is for this reason sometimes called a four-leaved rose. * The author ha» designed plotting paper for polar oo({rdlnates on which concentric circles and radial lines are drawn in faint blue ink. This paper is desirable for the rapid and accurate plotting of polar cttrres. Published by the Yale Cooperative Corporation, New Hayen, Conn. 270 DIFFERENTIAL CALCULUS Trace the following curves. 1. ya(2a-x) = «». 2. (x2 + 4aa)y = 8a«. 3. ai^ = x«-6x«. 4. (y-x)2 = x«. 6. x' + y' = a*. 6. aJV = (6* - y*) (a + y)« 7. y = log X. 8. y = «-**. 10..y=:(xa-l)a. 11. y = Binx. 12. y = tan x. 13. x»(y - o) = a' - xy«. 14. x*-2ax»y-axy« + aV = 16. (x2 + y«)» = aa{as«-y«). 16. p = a cos 2 e. 17. p = a Bin 3 ^. 18. p = a (1 — cos &). 19. p = asin'-. 3 20. p = a8ec>-. 3 CHAPTER XXII APPLICATIONS TO GEOMETRY OF SPACE tA«»v-fAy,s-fAx) 176. Tangent line and normal plane to a skew curve whose equa- tions are given in parametric form. The student is already familiar with the parametric representation of a plane curve. To extend this notion to curves in space, let the coordinates of any point (x, y, z) on a skew curve be given as functions of some fourth variable t, thus, (A) x^4>(t), !/=ir{t), z^xi^y The elimination of the parameter t between these equations two by two gives the projecting cylinders of the curve on the coordinate planes. Let the point F (x^ y, z) correspond to the value t of the parameter, and the point F'{x + Ax, y + Ay, z + As) correspond to the value f + A^; t^ Ay, As being the increments oix^y^z due to the increment Af as found from equations (A). From Analytic Geometry we know that the direction cosines of the secant (diagonal) FF^ are proportional to At, Ay, As;; or, dividing through by A^ and denoting the direction angles of the secant by a\ /8', 7', ^, , At Ay As cos a' : cos /8' : cos 7' : : -— ■ : -77 : -— • A^ A^ A^ Now let P' approach F along the curve. Then Ae, and therefore also Aa:, Ay, A«, will approach zero as a limit, the secant FF^ will approach the tangent line to the curve at P as a limiting position, and we shall have _ dx ay dz cos a : cos p : cos 7 : : — : -^ : — » 271 272 DIFFERENTIAL CALCULUS where a, /8, 7 are the direction angles of the tangent (or curve) at P. Hence the equations of the tangent line to the curve at the point (Zj y, z) are given hy (67) <fag dy dz ' dt dt dt and the equMion of the normal plane, i.e. the plane passing through (a:, y, z) perpendicular to the tangent, is m g(X-a:) + g(r-y) + g(Z-z)=:0, X, Yy Z being the variable coordinates. Ex. 1. Find the equations of the tangent and the equation of the normal plane to the helix* (6 being the parameter), X = acos^, ^ = a sin ^, (a) at any point ; (b) when 9 = 2 r. SohUion, dz . - dy ^ dz ^ = — a sm ^ = — y, -^ = a cos = x. — = 6. d$ de de Substituting in (67) and (68), we get at (x, y^ z), X^x T-y Z-z ^ ,„ = = — : — » tangent line ; — y X b and - y (X - x) + x(I^- V) + b{Z - z) = 0, normal plane. When tf =2 r, the point of contact is (a, 0, 2 bw), Pvi"« X-a _ T-0 __ Z-2bK " a " b ' « or, X = a, 6F = aZ — 2 abx, the equations of the tangent line ; and aF-fftZ -26aT = 0, the equation of the normal plane. * The helix may be defined as a cunre traced on a right oiroular cylinder so as to cut all the elements at the same angle. Take QZ as the axis of the cylinder and the point of starting in OX at Pq, Let aaradlus of base of cylinder and wangle of rotation. By definition, PN PK z , ,, ^ ■ ""c^r = :r 7rT-= "^ = *' (const.), or , z = ake. Sn aro P„y a6 Let aA-=» & ; then z = b0. Also, y = xMX= a sin 0,x- OM ~ a cos A APPLICATIONS TO GEOMETRY OF SPACE 273 177. Tangent plane to a surface. A straight line is said to be tangent to a surface at a point F if it is the limiting position of a secant through F and a neighboring point P' on the surface, when P' is made to approach P along the surface. We now proceed to establish a theorem of fundamental importance. Theorem. All tangent lines to a surface at a given point * lie in general in a plane called the tangent plane at that point. Proof Let {A) F{x,y,z)^0 be the equation of the given surface, and let F{x^ y, z) be the given point on the surface. If now P' be made to approach P along a curve C lying on the surface and passing through P and P', then evidently the secant FF^ approaches the position of a tangent to the curve C at P. Now let the equations of the curve C be {B) x=.4>{t), y = ir(t), ^ = X(0. Then the equation (A) must be satisfied identically by these values, and since the total differential of (A) when Xy y, z are defined by {B) must vanish, we have (C) 7-H f--\ r = 0- (50), p. 199 ^ ^ dx dt dy dt dz dt ^ ^' ^ This equation shows that the tangent line to C, whose direction cosines are proportional to dx dy dz It' w w p- ^^^ is perpendicular f to a line whose direction cosines are determined by the ratios dF dF dF ^_- • —— * » dx' dy' dz and since C is any curve on the surface through P, it follows at • The point in question is assumed to be an ordinary (non-singular) point of the surface, i.e. fijfi ^w ^f -—«—-»—- are not all zero at the point. ox cy dz t From Solid Analytic Geometry ve know that if two lines haying the direction cosines oosai, cot^xi <i08Yi and cos a,, cos^t, cosy, are perpendicular, then cos Ox cos oj + cos Px cos /3t + cos 71 cos 7,= 0. 274 DIFFERENTIAL CALCULUS once, if we replace the point P {x^ y, z) by F^ (a^j, y^, z^^ that all tangent lines to the surface at P^ lie in the plane* which is then the forrmdafor finding the equation of a plane tangent at (oTj, y^, z^ to a surface whose equation is given in the form F{x,y,z)^0. In case the equation of the surface is given in the form z =/(a:, y), let (2>) P(a:,y, z)=/(2:, y)-z = 0. Th n dF^d£^dz_ dF^^^dz ^^_^ dx dx dx cy dy dy dz If we evaluate these at {xy, y^, z^ and substitute in (69), we get which is then the formula for finding the equation of a plane tangent at (Xj, y^ 2j) io a surface whose equation is given in the form z=if(Xj y). In § 137, p. 200, we found (53), the total differential of a function u (or z) of x and V, namely, (E) dz = — cte H — dy. ^ ' dz ty We have now a means of interpreting this result geometrically. For the tangent plane to the surface z =/(x, y) at (x, y, z) is, from (70), (F) Z-2 = ^(X-x) + J?(r-y), oz cy Xy F, Z denoting the variable co-ordinates of any point on the plane. If we substitute jr = * + dxandr=» + (ly in (F), there results (G) Z'-z = ^dz + ^dy. ^ ' dz dy ^ Comparing {JE) and (G^, we get (H) dx^Z-z, Hence • The direction coeines of the normal to the plane (69) are proportional to ^*, ^i ^. dX) oyx dzi Hence from Analytic Geometry we see that (C) is the condition that the tangents whose direc- tion cosines are cos a, cos^, cos y are perpendicular to the normal ; i.e. the tangents must lie in the plane. t In agreement with our former practice, dFi dF^ dF^ dz^ dz^ d^x dvi C2, dxj dpx denote the values of the partial derivatiTes at the point (x^, y^, s,). APPLICATIONS TO GEOMETRY OF SPACE 275 The total differential of a function f{x, t/) corretponding to the incre- menit dz and dp equtdi the eorreipondin^ inerement (if the t eoordtnate (>f the (anient plane to the tur/ace t =f(z, j/). Thus, In the figure, PP" U the pl&ne tangent t« nuiace PQ &t P(x, y, i). Let AB = dx and CD = dj/; then di = Z-t = DP"- DE = EP". 178. Nonnal line to a sorface. The normal line to a surface at a given point is the line passing through the point perpendicular to the tangent plane to the surface at that point. The direction cosines of any line perpendicular to the tangent plane (A9) are proportional to e/\ £^ e^ ^ ' dF\ BFx BF\ dxi dffx &*i. are the equationt of the normal line • to the tnrface F{x, y, z) = at Similarly, from (J#), t>xi dyi are the equatwnt of the normal line* to the surface Zi=f{x, y) at • See lootnole. p. ZT4. 276 DIFFERENTIAL CALCULUS EXAMPLES 1. Find the equation of the tangent plane and the equations of the normal line to the sphere x> + 2^ + 2^ = 14 at the point (1, 2, 3). Solution. Let F{x, y, «) = x» + y^ + «« - 14 ; dF dF dF then — = 2x, — = 2 y, — = 2 z ; Xi = 1, yi = 2, «i = 3. dz dy dz dxi dyi dzi Substituting in (60), 2(x- 1) + 4(y- 2) + 6(z -3) = 0, x + 2y + 3z = 14, the tangent plane. X— 1 y-2 z-3 SubsUtuting in (71), '» 2 4 6 giving z = 3 X and 2 z = 3 y, equations of the normal line. 2. Find thd equation of the tangent plane and the equations of the normal line to the ellipsoid 4x> + 9y9 -f- 36 z^ = 36, at point of contact where x = 2, y = 1, and z is positive. ^n«. Tangent plane, 8(x - 2) + 9(y - 1) + 6 Vll(z - J Vil) = 0, ... x-2 y-1 z-jVu normal hne, = ~ = =-==— * 8 9 6vTl 3. Find the equation of the tangent plane to the elliptic paraboloid z = 2 x> + 4 y* at the point (2, 1, 12). Am. 8x + 8y-z = 12. 4. Find the equations of the normal line to the hyperboloid of one sheet «" - 4y» + 2z2 = 6 at (2, 2, 3). Ana. y + 4x = 10, 3x - z = 3. 5. Find the equation of the tangent plane to the hyperboloid of two sheets 6. Find the equation of the tangent plane at the point (Xi, yi, zi) on the surface ckb' + 6y* -f- cz« + d = 0. Ans. axiX + byiy + cziZ + d = 0. 7. Show that the equation of the plane tangent to the sphere x* + y' + 2* + 2Lx-f2 3fy + 2^z + D = at the point (xi, yi, Zi) is xix + yiy + 2i2 + i(x + Xi) + M(y + vi) + N(z + 21) + i) = 0. 8. Find the equation of the tangent plane at any point of the surface «' + y' + 2' = a', and show that the sum of the squares of the intercepts on the axes made by the tangent plane is constant. 9. Prove that the tetrahedron formed by the co5rdinate planes and any tangent plane to the surface xyz = a^ is of constant volume. APPLICATIONS TO GEOMETRY OF SPACE 277 179. Another form of the equations of the tangent line to a skew curve. If the curve in question be the curve of intersection AB of the two surfaces F(xy y, 2) = and G (2;, y, z) == 0, the tangent line FT at JP(ari, y^, z^) is the intersection of the tangent planes CD and CF at that point, for it is also tangent to both surfaces and hence must lie in both tangent planes. The equations of the two tangent planes at F are, from («9), fdFx, V , dFx, . , (a? — Xx) + -r—(y - Vi) + (78) dxx dOx dyx dFx dzx (z - zx) = O, (05 - 051) + ^\y -yi) + ^(Z- Zx) = O. dxx ^ '' ' dyx ^^ "" ' dzx Taken simultaneously, the equations (78) are the eqiuitians of the tangent line FT to the skew curve AB, Equations (78) in more compact form are 35 — 051 y — yi « — Zx (M) dFxdGx dFxdGx dFxdGx dFxdGx dFxdOx dFxdOx dyx dzx dzx dyx dzx ctei dxx dzx dxx dyx dyx dxx or, using determinants, 05 — 051 (W _ y — yi _ as — gj dFx dFx dyx dzx dGxdOx dyx dzx dFx dFx dzx dxx dGxdOx dzx dxx dFx dFx dxx dyx dOxdGx dxx dyx 180. Another form of the equation of the normal plane to a skew curve. The normal plane to a skew curve at a given point has already been defined as the plane passing through that point per- pendicular to the tangent line to the curve at that point. Thus, in the above figure, PHI is the normal plane to the curve AB at P. Since this plane is perpendicular to (75), we have at once. (W dFx dFx dyx dzx dGxdOx dyx dzx (X — 05i) + dFx dFx dZx dxx dGx dGx dzx dxx (y - vi) + dFx dFx dxx dyx dGx dGx dxx dyx (z - Zx) = O, the equation of the normal plane to a skew curve. DIFFERENTIAL CALCULUS BZAKPLE3 1. Find the eqafttiooa of the tangent Une and the equation o( the normal plane at (r, r, r W) to the curve of intersection of the sphere and cjlinder whose equa- tions are reapecUvel; ^ + j/* + z' = 4t', x* + v' = 2tx. Solution. LetF=i' + v* + «'-4r' and O^xf + y'-in. dXi dj/i dXi ?°! = 0, ?°! = 2r, ?2! = 0. SX) tgi dti Substituting in (76), ^-^ = '^^^ = ^^^ ; — ^2 1 or, p = r,a + v'2j = ar, the equations of the tangent PT at P to the curve of intersection. Substituting in (76), we get the equation of the normal plane, - v^ (I - r) + (1/ - r) + (I - r Ve) = 0, or, V2x-z = 0. 2. Find the equations of the tangent line to the circle »' + y* + *" = 26, a + I = 5, at the point (2, 2 Vs, 3). ^^ 2^ + 2 Vl:, + 8r = 26, . + . = & 3. Find the equation of normal plane to the curve ji + ^ + iS = ra, iS _ n + yi = 0, "tfJi, 1/1, Zi). Ana. 2yiz,x - {2xi ~ r)z,v - rtfit = 0. APPLICATIONS TO GEOMETRY OF SPACE 279 4. The equations of a helix (spiral) are «2 + y« = r«, y = xtan-* c Show that at the point (asi, yu Zi) the equations of the tangent line are c(x-xi)-f-yi(z-«i) = 0, c(y-yi)-xi(«-2i) = 0; and the equation of the normal plane is ViX — Xiv — c (z — «i) =s 0. x' t/* z^ 5. A skew curve is formed by the intersection of the cone — + tt — :; = ' a^ t^ c^ and the sphere x^ + ^ + a^ = r^. Show that at the point (xi, yi, Zi) the equations of the tangent line to the curve are c^{a^ - 6*)xi(x ^ xi) = - a»(6a + c«)zi(z - Zi), c'^ia^ - 62) yi (y - yi) = + 6»(c« + aa)zi (z - Zi) ; and the equation of the normal plane is a»(6» + c») yiZiX - 6^ (c« + a^) ziXiy - c^ (a* - 6^ XiViZ = 0. CHAPTER XXin CURVES FOR R£F£R£irC£ For the conyenience of the student a number of the more common curves employed in the text are collected here. Cubical Parabola Y Semicubical Parabola y = ax*. t^ = ax\ The Witch of Aqnesi r 1 -^K 'J ^"^^ o X xhj=z4^a^(2a^y). The Cissoid of Diocles y^(2a — x) — x\' p = 2 a sin tan 9. 280 CURVES FOR REFERENCE 281 The Lemniscate of Bebnoulli The Conchoid of Nigomedes. p* = a* cos 2 $. p = a sec $:tb. Cycloid, Ordinary Case Cycloid, Vertex at Origin Y x=:a arc vers {:: ^-V2ay-y». a a (^ — sin 6), a(l — cos^. X = t: a arc vers - -h '>J2ay a a (^ + sin ^), a(l — cos fl). -y^ Parabola y=5(^ + ^"")- as* + y* = a*. 1 282 DIFFERENTIAL CALCULUS Hypocycloid of Fouk Cusps EVOLUTB OF ElLIPSB 05* -f y* = a*. ' x=za cos'^, (ooj)* + (hy)^ = (a" - h^K Cakdioid Folium of Descabtes 35' -f y' + «a5 = a Vx* -f y\ p = a (1 — cos ^). «' -h y* — 3 oay = 0. CURVES FOR REFERENCE 283 L1MA90N p=zh — a cos 0. yl J— X* Spiral of Asghimedss Logarithmic or Equiangular Spiral = aft p = «*»•, or log p = a0. Hyperbolic or Reciprocal Spiral L1TUU8 pO=ia. pV = a*. 284 DIFFERENTIAL CALCULUS Fababolig Spiral Logarithmic Curve ^, -" (p--a)" = 4a(j^. y = log X. Exponential Curve Probability Curve y = «'. y = e"*^. Sine Curve Tangent Curve o y = 8in X, y = tana;. CURVES FOR REFERENCE 285 Thbee-Leaved Rose Three-Leaved Rose p =za sin 3 $, p = a cos 3 $, Four-Leaved Rose Four-Leaved Rose p = a sin 2 $. p = a cos 2 B. INTEGRAL CALCULUS CHAPTER XXIV INTEGRATIOir. RULES FOR DrTEGRATDrO STANDARD ELEMENTART FORMS 18L Integration. The student is already familiar with the mutuallj inverse operations of addition and subtraction, multipli- cation and division, involution and evolution. In the examples which follow the second members of one column are respectively the inverse of the second members of the other column. y = a? + l, a; = ± Vy-1; y zzzamccj x=z arc sin y. From the Differential Calculus we have learned how to calcu- late the derivative f{x) of a given function /(x), an operation indicated by ^ or, if we are using differentials, by df(x)=:f{x)dx. The problems of the Integral Calculus depend on the mver%e operation^ namely : To find a function f{x) whose derivative (A) f(x) = 4>{x) is given. Or, since it is customary to use differentials in the Integral Calculus, we may write (B) df{x):=.f(x)dx=^<l>{x)dxy and state the problem as follows : Saving given the differential of a function^ find the function itself, 287 288 INTEGRAL CALCULUS The function /(a;) thus found is called an integral* of the given differential expression, the process of finding it is called integratwn^ and the operation is indicated by writing the integral sign t j in front of the given differential expression. Thus, (O) ff{x)dxt^f{x), read, an integral of f{x)dx equals f{x). The differential dx indi- cates that X is the variable of integration. For example, (a) If f{x) = a?y then f(x)dx=^^ a^cfa, and (b) If f{x) = sin x, then f (x) dx = cos xdx^ and / cos xdx = sin x. dx (c) If /(a:) = arc tan a:, then f(x)dx==- — -^» and dx A = arc tan x. +7? Let us now emphasize what is apparent from the preceding explanations, namely, that Differentiation and integration are inverse operations. Differentiating (O) gives (2>) dCf (x) dx =f (x) dx. Substituting the value oif{x) dx[=z (if{x)] from (B) in ((7), we get (E) jdf{x)^f{x). d C Therefore, considered as symbols of operation, — and I • . . <2a; are inverse to each other; or, if we are using differentials, d and I are inverse to each other. * Called antv-d\fftreniiaX by some writers. t Historically this sign is a distorted <9, the initial letter of the word turn. Instead of defining integration as the inrerse of differentiation we may define it as a process of summation, a rery Important notion which we will consider In Cliapter XXX. X Some authors write this U^ V(^) when they wish to emphasize the fact that it is an inverse operation. INTEGRATION 289 When d is followed by j they annul each other, as in (D), but when j is followed by (2, as in (E)^ that will not in general be the case' unless we ignore the comtanb of integration. The reason for this will appear at once from the definition of the constant of integration given in the next section. 182. Constant of Integration. Indefinite integral. From the pre- ceding section it follows that since d(Qi?) =3 a^dx^ we have | 8 a^dx = a? ; since d(2? -|- 2) = 8 a^do^ we have j 3 3?dx = a?" -f- 2 ; since d(aj" — 7) = 8 st^dx^ we have j 3 a^dx = a? — 1. In fact, since d(a:»+(7)=3a?(ir, where C is any arbitrary constant, we have r3a?(ic = 2^-f C. A constant C arising in this way is called a constant of integration* Since we can give C as many values as we please, it follows that if a given differential expression has one integral, it has infinitely many differing only by constants. Hence /^ f(x)dx=f{x)+C; and since C is unknown and indefinite, the expression f{x) + c is called the indefinite integral off{x) dx. It is evident that if ^ (x) is a function the derivative of which is f(x)y then ^ (x) -f C, where C is any constant whatever, is likewise a function the derivative of which isf{x). Hence the Theorem. If two functions differ by a constant, they have the same derivative. • ConstMit here means that it is Independent of the variahle of integration. 290 INTEGRAL CALCULUS It is, however, not obvious that if ^ (2:) is a function the deriva- tive of which is/(2;), then all functions having the same derivative f(x) are of the form where C is any constant. In other words, there remains to be proven the Converse Theorem. If two functions have the same derivative^ their difference is a constant Proof Let ^(x) and -^ (x) be two functions having the common derivative f(x) . Place F{x) = ^(x)— ylr(x); then (A) F' (x) =-§-[<l>{x)^it (x)] =f{x) -/(x) = 0. By hypothesis But from the Theorem of Mean Value, (44), p. 168, we have F{x^Ax)^F{x)==AxF'(x + 0'Ax). 0<tf<l .-. F{x + Ax)-F{x) = 0, [Since by (A) the derivative of F(x) ]a zero for all values of x.] and F(x + Ax) = F(x). This means that the function F{x)=.<l>(x)^ir(x) does not change in value at all when x takes on the increment Ax^ i.e. <f>{x) and '^{x) differ only by a constant. In any given case the value of C can be found when we know the value of the integral for some value of the variable, and this will be illustrated by numerous examples in the next chapter. For the present we shall content ourselves with first learning how to find the indefinite integrals of given differential expressions. In what follows we shall assume that every continuous functi(m has an indefinite integral^ a statement the rigorous proof of which is beyond the scope of this book. For all elementary functions, however, the truth of the statement will appear in the chapters which follow. In all cases of indefinite integration the test to be applied in verifying the results is that the differential of the integral must he equal to the given differential expression. INTEGRATION 291 183. Rules for integrating standard elementary forms. The Differential Calculus furnished us with a General Rule for differ- entiation (p. 42). The Integral Calculus gives us no corresponding general rule that can be readily applied in practice for performing the inverse operation of integration.* Each case requires special treatment and we arrive at the integral of a given differential expression through our previous knowledge of the known results of differentiation. That is, we must be able to answer the question, What function^ when differentiated^ mil yield the given differential expression f Integration then is essentially a tentative process, and to expedite the work, tables of known integrals are formed called standard forms. To effect any integration we compare the given differential expression with these forms, and if it is found to be identical with one of them, the integral is known. If it is not iden- tical with one of them, we strive to reduce it to one of the standard forms by various methods, many of which employ artifices which can be suggested by practice only. Accordingly a large portion of our treatise on the Integral Calculus will be devoted to the explanation of methods for integrating those functions which fre- quently appear in the process of solving practical problems. From any result of differentiation may always be derived a formula for integration. The following two rules are useful in reducing differential expressions to standard forms. (a) The integral of any algebraic sum of differential expressions equals the same algebraic sum of the integrals of these expressions taken separately. , Proof Differentiating the expression \ du-\' \ dv — J dw^ w, v, w being functions of a single variable, we get du-^dv — dw. III, p. 144 [1] .-. i(du -f dv — dw) = t du + i dv ^ j dw. * Eren though the Integral of a glren differential expression may be known to exist, yet It may not be possible for us actually to find it in terms of known functions, because there are functions other than the elementary functions whose deriyatives are elementary functions. 292 INTEGRAL CALCULUS (6) A constant factor may be written either btfore or after the integral sign. Proof Differentiating the expression a i dv gives adv, IV, p. 144 [2] •*• I adv = a I dv. On account of their importance we shall write the above two rules as formulas at the head of the following list of STANDARD ELfiMENTART FORMS [1] . j (du + cfv — dw) = I du + Idv — j dw. [9] i adv = a I dv, [8] Cdx = a? + C. [4] Cv**dv = ^^^ + C. n=^-l [« [10] n+1 [5] J^:=logV+C = logr V + log c = logr cr. [Placing C^logc] [6] Ca^dv = -^5l_ + c. J log a [»] fe^'cli; = c« + C. [8] r sin vdv = '- cos v + C- I cos vdt; = sin v + C I sec* vdr = tan v + C. [11] I cosec* v ffv = — cot V + €• [19] r secf tanf dt; = secf + C. INTEGRATION 293 [18] j C8C V cot t» «f V = — CSC V + C* [14] Ctaxkvdv = log sect; + C. [15] I cot vdv = log sin v + C [16] J sec vdv = logrtan (h + t) + ^• [IT] fcsc vdv = log taH (^) + C. J t;» + a* a a [18] [19] [20] f ,^^ = arc sin ^ + C. [21] C—M= = log (V + Vv* ± a«) + e. , = arc vers- + C. V2av - V* « — ^ = — arc sec— + C [28] Proof 0/ [8]. Since rf(a: + C)=<fo, II, p. 144 we get j dx=:x + C. Proof of [4]. Since d (J—- + C^ = ifdv, VII, p. 144 \«-hl / J^^n + l trdv = 1"*"^* This holds true for all values of n except w = — 1. For, when » = — 1, [4] gives v''dv = Ai T + C = - + C=oo + C7, — 1 + 1 which has no meaning. The case when n = — 1 comes under [5]. 294 INTEGRAL CALCULUS Proof of [5]. Since d(logt; + C) = -^ IX, p. 145 we get r — = log v + C. The results we get from [5] may be put in more compact form if we denote the constant of integration by log c. Thus, I — = log V + log c = log cv. Formula [5] states that if the eapression under the integral ngn i$ a fraction whose numerator is the differential of the denominator, then the integral is the natural logarithm of the denominator. BXAMPLBS For formulas [l]-[6]. Verify the following integrations. sfidx = — H C, by [4], where v = x and n = 6. 2. faz^dx = aCz^dx by [2] = — + 0. By [4] 3. J{2s^-6x*---Sx-^4t)dx=j22*dz-j6x^»dt-'Csxdx-^C4dz by [1] = 2 fx^dr - bCx^dx - sCxdx + iCdx by [2] = 2-X--i- + '^-^^- Note. Although each separate Integration reqalres an arbitrary constant, we write down only a single constant denoting their algebraic sum. 4. J(-^-^ + 3cVS2^dx=J'2ax-*dx- ffcx-'dxH- r8cx«dx by [1] = 2 ajx-^dx - bCx-^dx + 8 cCx^dx by [2] = 2a.?^-5.^ + ?^ + C l^[4] = 4aVx + -+fcx* + C. X 6. j2ax*-i(ix = ^ + C. 6. Js m^^dz = ?^ + C. INTEGRATION 295 JBinl. First eziMmd. (nx) • 13. /(a. + M«V«ix = <^±^ + C. Hint, ThiB may be brought to form [4]. For let v^a'+Mz* and «»); then <iv»2M«df. If we now insert the constant factor 26* before xdx, and its reciprocal — before the integral sign (so as not to change the ralue of the expression), the expression may be integrated, using [4], namely, v^dv=- + C. n + 1 Thns, r(a« + 6»jB«)*«ite«-^ r(a»+6«a:^*26«a:d;t--^ r(a» + 6»JB«)*<l(a> + M«^ "26«' I '^^" 36« "*'^' JVbf«. The student is warned against transferring any function of the Tarlable from one tide of the integral sign to the other, since that would change the value of the integral. 14. fVa* - x^xdz = - J(a2 - x«)i + C. 15. C{Sax^ + 4te«)*(2(ix + 46a;«)dx = }(8axa + 4te«)* + C. Hint. Use [4], making va3ax*+46a;* and nB|. 16. r&(6axa + 86x»)«(2ax + 4ea«)dr = — (6ax« + 865e«)* + C. •/ 16 !''• f . .i = i(«' + g')* + <^- J (a« + x«)* J7i»<. Write this C(a^ + a*)" * afldx and apply [4]. 18. f ^^. = -2Vl-a-|.C. ^ Vl-x 19. j2ry(^ + l)*dy = |^(ya + p«)t + C. 296 INTEGRAL CALCULUS 20 f ^a^da ^ a Solution. Saxdx ^ r xdx ri^axax ^ r xax J 62 + c2x« J 62 + e^« ^^^ This resembles [5]. IVir let o«6*+e^*; then dv=2^xdx. If we introdnoe the faetor 2e* after the integral sign, and — - before it, we haye not changed the yalue of the expression, but the numerator is now seen to be the differential of the denominator. Therefore r Xdx ^8a/-2c^^3a/-d(62 + e^')^8a J 62 + 62x2 2d2j 62 + ^2 2e2J 62 + e2x2 2e2 ^^ ^ '^ ^^^ 22. r-|^ = llog(x2-l) + logC = logC Vx23i. •/ x* — 1 • -_ r 56xdx , c J8a-66x2 (8a-66x2)* 26. r-^ = x-^ + ^-log(x + l) + a. •/ X 4* 1 2 o ^«<. First divide the numerator by the denominator. 27. r?^^dx = x -log(2x + 8)2 + C. J 2x + 3 28. r^!lillldx = ilog(x«-nx) + C. /fti-icK 1 a + 6c* ?io 31. J (log a)8 ^ = J (log a)* + C. 32. t2 + 1 . f^— ti(fr = ^ + r + 21og(r-l) + a. J r — 1 2 J (a + 6M»)"» 6n(l — m) Proofs of [6] and [1]. These follow at once from the corre- sponding formula for differentiation, X and Xa, p. 145. INTEGRATION 297 For formulas [6] and [7]. Verify the following integrations. J 2 log a 8oi.Miicm, Cb(fi*dx = bfa^'dx. By [2] Tbla reeembles [6]. Let r b2x ; then dv'^2dx. If we then insert the factor 2 before dx and Jie factor \ before the integral sign, we hare 6/«..dx = ^/a.^<fc = 5ja.^d(2«) = ^.j^+C. B, [6] 4. J c"dte = ne» + C 6. r6^+**+«(x + 2) dx = ief^ + te + « + C. 6. ((a^-lF^)dx = -^ — r + C. J n log a m logo 7. (a'tFdx^—^^ + C. J 1 + log a (e« 4- e •)dx = a(€« - e «) + C. 11. J.^ = ^ + c- + log(c--l) + C. 12. •e^-1 r?— idr = log(e^ + 1)2 - r + C. J er + l 13. r ('"-f)' dx= °'^'^,°-t -2x+c. J a*&^ log a — log 6 Proofs of [8]-[18]. These follow at once from the corresponding formulas for differentiation, XII, etc., p. 145. 298. INTEGRAL CALCULUS sin vdv /•— sin vdv Proof of [14]. ftan vdv = C?^^ = _ f •^ •/ COS V J COS V / d (cos ty) COS V = -logcosv + C by [5] = log sec V -f- C [3i]ioe-logQOflVi--log "-lQgl + logaeoi7»loffieov.] Pro./ of [15]. foot ri« = f^^^ = f-^i^illi^ •^ •/ Sin V J sin v = log sin t; + C. By [5] Proof of [17]. Since esc t; = esc v CSC V — cot V CSC V — cot V •^ •/ CSC V — cot v _ (* d{(iSG y — coty) CSC V — cot V r = log (CSC v — cot v) -f. C by [5] , /I — cos v\ . ^ = log( — : ] + C \ sm t; y 2sin»^ =iog — ; — v'^^ ^y ^'^' P- ^' ^^' p- ^ 2 sin- cos - = logtan^ + C. Pr<?<?/ 0/ [16], Substituting v + ^ for v in [17] gives J'csc U + 1^ rft> = log tan ^1 + 1 V C. But cscf t; + ^ j = seow; therefore J aeo vdv = log tan /^| + 1'\ + c. INTEGRATION 299 For fornralas [8]-[17]. Verify the following integratiomu 1 r • o J coB2ax , ^ 1. I sm 2 axdx = h C, J , 2a Solution, This resembles [8]. For let v « 2 ox ; then dv « 2 ckte. If we now insert the factor 2a before dx and the factor — before the integral sign, we get /Bin 2 axdx = — I sin 2 ax* 2 adi 2aJ \ r 1 = — I Bin 2 ox -(2(2 ox) = cob 2 ox + C by [8] 2aJ ^ 2a *■ '' coB2aa; ^ 2a 2. fcoBmxdserr— Binnix + C 3. IdBec'tedx = -tan&G6 + C J m Jo 4. r^coB--8in8tfW = 38in- + }co88tf + a. 5. r7 8ec8atsn8ada = }8ec8a + C. ik cos (a + by)dy = ^sin (a + ^) + C 7. rcoaec«a5»-««dx = - Jcotx« + C. 8. r 4 C8C ox cot axdx = — esc ox + C^> •/ a . r Binxdx , c 9. I — = log J a + & COB X 1 (a + & cos x)h . re«»*Binxdx = -c«»* + C. ^' f «/^ _ =-^tan(a-6x) + C. J COS* (a — te) 2. fcosnogx)— = 8inlogx + C. 14. f <L±^?^5l^ = log (x + gin x) c. •/ . X •/ X -f Bin X 3. r-=^ = -ncot- + C. 16. I — rrr=:8ec0 + C. J ^j^,x n J co8>0 n 6. r (tan a + cot a)^da = tan a - cot o + C. 800 INTEGRAL CALCULUS 17. r(86C/9-tan/3)3d/3 = 2(tan/9-8ec/3)-/3 + C. ai+iintf COS ede = 4- C. log a 19. r(tan2u-l)*du = ^tan2u + logco82tt + C. «^ r. . . . vj log (1 — tan a tan y) , ^ 20. I tan y tan (y + a)cl2^ = - y ^^ ^ + C. J tana Proof of [18] . Since .. i ("- aro tan - + (7^ = - ^7 =-^, by XIV, p. 146 © rfv 1 . V we get r -r r = - arc tan - + C* Proof 0/ [19]. Since -5^, = ^ f-^^ —\ , cj^^:lc(j, ?_v. •/ v^ — c? 2aJ yv — a v-\-aJ = 2^11og(v-«)-log(t; + a)KC by [5] 2a v-f-a Pro^ of [20]. Since <-) rf/^arosin-H-cV . ^''^ ^ rft. ^ by XIX, p. 145 we ffet I — , = arc Bin - + (;• •Atoodl-arccot-+ C'|= — — — ; and I --—,= - -arc cot- + C Henoe - arc tan - + C=» — arc cot - + C. v* + a* a a a a Since arc tan - + arc cot - = - 1 we see that one result niay be easily transformed into the other. a a 2 \ ^ The same Icind of discussion may be given for [20] inyolying arc sin - and arc cos -i and for [23] InTolTing arc sec - and arc oc - . INTEGRATION 801 Proof of [21]. Assume Vt;*-f.a* = 2, a new variable. Then ^ -{-€?=. ^^ and differentiating, 2 vdv z=. 2 2k2a;, or, dv dz Z V By composition in proportion, dv dz dv 4- dz therefore Z V V -{-z dv dv + dz Vt^Tf^ v^-z [EeplaolBg s by Its equal V»«+o«.] Hence ^ r^^^^ /-rf^:^ = log(t; + g) +^ by [5] = log(v + Vv* + a*) + C. In the same way by assuming Vv* — a' = 2, we get f . / , = log(v + y/v'-a^ + C. •^ V v'* — a Proofs of [%i] and [38]. These follow at once from the corre- sponding^ formulas for differentiation, IXIII and IIY, p. 48. EXAMPLES For formulas [18]-[23]. Verify the following integrations. ^ r dx 1 2x ^ 1. I — - — - = - arc tan -— + C. -^ 4x«4-9 6 8 SolutUm, This resembles [18]. For, let v^z=4x^ and a< = 0; then v=:2x, do = 2 dx, and a = 8. Hence if we multiply the numerator by 2 and divide in front of the integral sign by 2, we get / dx __ 1 r 2dx _ 1 r d{2x ) 4x2 4-9 " 2 J (2x)« -f (3)2 " 2 J (2x)2 + 1 2x (3) 2 =r - arc tan —- + C. By [18] 6 8 802 INTEGRAL CALCULUS 6. f_jL= = hog(6y+V^3;^ + a. 6. I _ = - arc am z« + C 8. I — , = - arc aec •— + C •^ Vl31? 2 -^ xV4x«-9 3 8 7 f — — - arc tan — 4- C 9. I . = arc vers - + C. ^ r edt e, W + a.^ /• 7d» 7 . ^/6 , ^ 1. I , z = —^Mcsm\- 8-^-0, , /• coBcKia 1 . /8ina\ , ^ ^- I -s TT- = - arc tani ) + C. I. r = arc sin c* + C >. I — ^ = arc sin (log x) + C •^ « Vl - logsx 6. I ^ = arc sin \- C. •^ Vaa - (u + 6)« « _ f adz a ^ 2- c , _ 7. I — r- = 7 arc tan — h C, ^ r dx 1 x + l^ 8- |-^ — ^ ^= arctan-^ + C. J x2 + 2x + 5 2 2 Ainf. By oompletiog the square In the denominator this expression may be brought to a form similar to that of Ex. 17. Thus, -si =z \ »-arctan + C. By [181 a4 + 2j: + 6 J (x« + 2x + l)+4 J(x+l)« + 4 2 2 '^ ^ Here 9 » a; + 1 and a* 2. /nn^ Bring this to the form of Ex. 16 by completing the square. I!hu8, /dx r dx r dx C dx . 2x-l ^ _ ^^, , "1 . =) , ^ ==\- -arcsin— -— +C. By [20] X9. I — = arc sin — \- C. Here v = x-^ and a« f . 20. f ^ =-4arctan^^ + C, J 1 + X + x^ INTEGRATION 803 dz 21. r = = arcBin(2g-3) + (7. 23. f ^y =2-iog^y + «-^^a 26. fr-r — T - = arctan(2«-l) + a 26. f ^ : = log(< + a-f V2cM-H«)-f C> •^ V2a« + ^ €Wf C ^ I CX . ^ 27. I — 7=== = — axx5 8ec-7 + C. •^ X Vc^c* - a^fra act ao 28. r-4^= = iarcyerBl8x« + C. •^ Vx«-9x« 3 •/ a^ + X" a a 2 30. r^^"^dx = llog(3x»-2)--^log^"^-^ + C. 184. Trigonometric differentials. We shall now consider some trigonometric differentials of frequent occurrence which may be readily integrated by being transformed into standard forms by means of simple trigonometric reductions. Example I. To find I sin"* a; no^xdx. When either m or n is a positive odd integer, no matter what the other may be, this integration may be performed by means of formula [4], ^+i I ifdv = -' J n + l For the integral is reducible to the form I {terms involving only co%x)9inxdx^ when sin x has the odd exponent, and to the form I (terms involving only sin x) cos xdx^ 304 INTEGRAL CALCULUS when cos x has the odd exponent. We shall illustrate this by means of examples. Ex. 1. Find rsin^xcoB^xdas. iSoZution. fsin^x cos^ xdx = Isin^x cos^ x cob x(2x = fsin^x (1 - 8inSx)s cos xdx 28, p. 2 = r (sin^x — 2 Bin^x + sin^x) cos xdx = Msin x)> COB xdx — 2 Main x)^ cos x^x + j (sin ^^ cob xdx sin'x 2 8in^x Bin^x ^ « -„ , o o 7 Here o = Bin x, do = cob xdx, and n = 2, 4, and 6 respectively. Ex. 2. Find rco8*xdx. SoXvXvm. fcoB^xdx = j cos^x cos xdc = Ml — sin^x) cos xdx = jcos xdx — r Bin*x cob xdx sin^x . ^ = sin X — V C. EXAMPLES //* sin'x sin*x(2x = \ co8*x — cos x + 0. 2. I sin^x cob xdx = 1- C. 3. iBin^xdx = — cos x + - cob*x \- C. J 3 5 4. fcoB^xdx = sin X — - sin'x H h C. 5. I Bin « COB 808 = V C. 4/ 2 6. r coB^x sin'xdx = — J cosPx + } cos'^x + C. 7. ( COS*a sin ada = ^— ^ + C. •/ 3 Q fcos^xdx 1 « . /^ o rsin'oda . . ^ 8. 1—7-: — = C8Cx--CBC«x + C. 9. I — = 8eca + C06a + C. •/ sin^x 3 J cos^a 10. fsin^^ coB»0d^ = ^ sin^0 - J^ sin^^ + C. 11. fsin'^ co8«^d^ = { sin*d - ^^ am^e + V*? sin^*^ + C. INTEGRATION 805 12. f-^^dif = - 2 Vcos y (1 - ? coeay + -co8*y) + C. 13. r^?^ = ?8inlt(l-lBin8t + -8in<t) + C. V5^ 2 2 7 Example II. To find i tan* xdx, or I coV xdx. These forms can be readily integrated, when n is an integer, on somewhat the same plan as the previous examples. Ex. 1. Find ftan* xdx. BoLviwn, ftan* xdx = ftan^x (sec^ x - 1) dx 28, p. 2 = ftan^ X 8ec2 xdx — jtan* xdx = r(tan x)'cl(tan x) - ("(sec^x - l)(fe = — tan X + X + C. Example III. To find j 9e(f xdx^ or J cs(f xdx. These can be easily integrated when n is an even positive integer. Ex. 2. Find jsec^xdx. SohUUm. fsec^ xdx = r(tan«x + 1)* 8ec« xdx 28, p. 2 = Mtan x)* sec* xdx + 2 ntan x)' sec* xdx + Jsec* xdx = h2 htanx + C. 5 8 Example IV. To find ( tan'^x %e<f xdxj or i coi^x cs(f xdx. When n is a positive even integer we proceed as in Example III. Ex. 3. Find jtan^xsec^xcEx. SoMUm, fton^x Bec*xdx = ftan^x (tan^x + 1) sec^xdx 28, p. 2 = Otan x)* sec^xdx + Jtan^x aec'xdx ^tan^ tanTx^^ By [4] 9 7 Here « = tanx, do = sec* xdx, etc. 806 INTEGRAL CALCULUS When m is odd we may proceed as in the following example. Ex. 4. Find J tan^x sec'xda; = } tan^x sec'x sec x tan xdx = r(sec*x - 1)« sec^x sec x tan xdx 28, p. 2 = Hsec'x — 2 sec*x + sec«x) sec X tanxdsD sec'^x 28ec*x . sec^x ^ _ Here o = 8ecx, do = sec x tan xdx, etc. EXAMPLES 1. rtan»xdx = — ^- + logcosx + C. J 2 2. J'cot«xdx = -?^-log8inx + C. 3. fcot^^dx = ~ cot»5 + 3 cot? + X + C. •^8 S3 4. Jcot*xdx = -cotx-x + C. 6. fcot^oda = - } cot*« + i cot«o + log sin a + C. a ftan»^dy = tan*^ - 4 tan« 7 + 4 logsec ? + C. •^ 4 4 4 4 !• r . J tan^x 3tan'x 7. I 8ec»xdx = — — + ^^=^ + tan»x + tanx + a •^ 7 6 8. Jcsc^xdx = — cot X — } cot^x — \ cot* x + C 9. rtan«*Bec«^ = *55l* + *5B!^ + c. •^ 7 6 10. rtan*08ec*9({0 = |8ec'0 - J 8ec>0 + C. 12. Jtanlxi«c4«te = !l!^ + 2tan»«^p 14. P^ = tan. -2cot.- £2^ + 0. J tan*a 8 INTEGRATION 307 15. r(tan«z + tan*2)dz = ^tan»2 + C. 16. ("(tan t + cot «)8 (ft = J (tan« t - cot* <) + log tan^ t + C, Example V. To find isinTxcoif'xdx by means of multiple angles. When either m orn is a positive odd integer the shortest method is that shown in Example I, p. 303. When m and n are both positive even integers the given differential expression may be trans- formed by suitable trigonometric substitutions into an expression involving sines and cosines of multiple angles, and then integrated. For this purpose we employ the following formulas : sin ti cos u = i^ sin 2 u, 86, p. 2 sin*w = i — ^ cos 2 w, 38, p. 3 cos^te = ^ + ^ cos 2 u. 39, p. 3 Ex. 1. Find fcos^xdx. Solutum. Cco^xdx = f (i + J cos 2 x) dx 38, p. 3 = i fdz + - rco82«te=:- + 78in2x+C. 2J 2J 2 4 Ex. 2. Find fsinaxcos^xdx. Solution, fsin^xcosaxdx = J rflin22xdx 36, p. 2 = if a - i cos4x)(fe 38, p. 3 = sin 4 X + C 8 32 Ex. 3. Find fsin^xcos^xcte. Solution, r sin^ x cos^ xdx = j (sin x cos x)' sin^ xdx = rjsin22x(i-icos2x)dx 36, p. 2 ;. 38, p. 3 = J rsin32xdx - J rBin2 2xco8 2xdx = if a - ico8 4x)dx - J rsin«2xcos2x(ix _ X sin 4 X sin' 2x ^ ~ 16 64 48" 808 INTEGRAL CALCULUS Example VI. To find i sin mx cos nxdxj Csin mx sin nxdzj or Ccos mx cos nxdxj when m ^ n. By 41, p. 8, smmxcosnx = ^8iii(m + n)x-\'^Bm{m'-'n)x. .'. j sin mx cos nxdx = J^ rsin {m -f- n) scdx + J j sin (w — n) xdx _ cos(^+yt)^ C0B{m — n)x ^ ^ 2(w + n) 2{m — n) Similarly we find /, 8in(wi-|-n)a: . sin (m — n)a; . ^ sm mx sin nxdx = rr^ — ■ — ^ H — ^ '— + C, 2 (w -H n) 2 (m — n) J, 8in(7n-f-n)a: sin(m — n)a; . ^ cos WW? cos nxdx = — ;r-^ — ^ — ^— + -—-^ ^ + C 2(?n + n) ^ 2(w-n) EXAMPLES /x 1 co8«xdx = - + -8in2x + C. 2 4 2 3x sin 22 . sin 42 4. Isin^xdx = — ( 52 — 4 8in22 + — h-Bin42) + C •/ 16 \ 3 4/ 6*. rco8«2d2 = — (52 + 48in22 — -? + 78in42 ) + C. •/ 16 \ 3 4/ y. r . -1 o J 8in«2a .a 8ln4a _ 6. I Bin* a C08» odo = -— + + C. J 48 16 64 7 rsm*«coB«W< = — (3«-8in4« + ^^^) + C. •/ 128 V 8 / 8. rc(M^2 8in*2d2 = — (62 H- -8in«22 — 8in42 l + C J . 128 V 3 8 / A r o • e J cosSy C082y . _ 9. I COB 3y sin 5^(2^ = + C •i/\ r • c • a J sinllz sinz . ^ 10. I Bin 6 z Bin 6 zdz = \- h C. J 22 2 •■I r .1 ^ J sinlls . 8in3« . ^ 11. I cos 4 « cos 7 ad« = h C J 22 6 12. j COS} 2 sin} 2(22 = — I cos 2 + 006^2 + C- 13. Jcos 3 2 COS I xdx = ^\ sin Y 2 + A 8in J 2 + C. CHAPTER XXV COHSTAHT OF IHTXORAIIOII 185. Determination of the constant of integration hy means of initial conditions. As was pointed out on p. 290, the constant of integration may be found in any given case when we know the value of the integral for some value of the variable. In fact it is necessary, in order to be able to determine the constant of inte- gration, to have some data given in addition to the differential , expression to be integrated. Let us illustrate this by means of an example. Ex. 1. I^ad a. fonctioti whose flrat derivatiTs is SsC — 2z + 6, and which shall hare the ralne 12 when z = 1. Solution. (3 x^ — 2 z -)- 6) de 18 the diSerential expreaalou to be Integrated. Thus, r(3i» - 2i + 6)<it = x» - z» + 6* + C, where C is the constant of integration. From the conditlone of our problem this nsolt must equal 12 when z = 1 ; that is, 12 = 1-1 + 6 + 0, or, C = 7. Hence z*— z* + 6x + 7l8the required function. 186. Geometrical signification of tlie constant of integration. We shall illustrate this by means of examples. Ex. 1. Determine the equation of the curve at every point of which the tangent has the slope 2z. SobitUm. Since the slope of the tangent to a curve at any point Is — , we have by hypothesis *** dx or, in = 2xdx. Intagratlng, j/ = ifxdx,OT, (A) K = *'+C, where C is the constant of integration. Now if we give a series of values, say 0, 0, — 3, (A) yields ibe equations V = z» + 6, i/ = x\ v = 810 INTEGRAL CALCULUS irhoge loci are parabolas with axes coinciding with the sxia of y and having 6, 0, — S respecilvelf as inlercepte on the axis of Y. All of the parabolas (A) (there are an Infinite number of them) have the same value of — ; that is, they have the same direction (or slope) for the same value of X. It vrill also tie noticed that the difference In the lengths of their ordinates remains the same for all values of z. Hence all the parabolas can be obtained b; moving any one of them vertically up or down, the valae of C in this case not affecting the slope of the carve. If In the above example we impose the additional condition that the curve shall pass through the point (1, 4), then the coBrdinates of this point most satis:^ {A), *"°« 4.1 + C, or, C = 8. Hence the partlcnlat curve required Is the parabola y = a* + 3. Ex. 2. Determine the equation of a curve such that the slope of the tangent to the curve at any point Is the negative ratio of the abscissa to the ordinate. Solulion. The condition of the problem is ez* pressed by the equation • or, separating the variables, ydii = '-zdx. Integrating. ^ = -_ + c, or, I* + v« = 2 C. Tliis we see represents a series of concentric circlM with their centers at the origin. If, in addition, we impose the condition that the curve must pass through the point (3, 4), lb.n » + 16-20. Hence the particular curve required Is the circle x* + ^ = 26. 187. Physical signification of the constant of integration. The following examples will illustrate what is meant. s In a ett^ght Solution. Since the acceleration \=— from (U), p. lOfil ia constant, say/, h«vB L dt J dt ■^' ir, do =/d(. Integrating, (.i) e =^ + C. To delwmlne C, suppose that the initial velocity be «( r = Co when f = 0. CONSTANT OF INTEGRATION 811 These values substituted in {A) give to = + C, or, C = Vq. Hence (A) becomes (B) «=yi + «o. Since » = ;^ Fw, P. lOs] , we get from (B) or, da=ftdt + V(4t. Integrating, (C) «=iyr»-f ro« + C. To determine C, suppose that the initial space (= distance) be «o; that is, let s = So when t = 0. These values substituted in (C) give «o = -f + C, or, C = So. £ence (C) becomes (D) » = i/P + »o« + »o. By substituting the values f=g, Oo = 0, So = 0, s = ft in (£) and (D) we get the laws of motion of a body falling from rest in a vacuum, namely, ' (Ba) v — gt^ and (Da) A = ip^. Eliminating t between (Ba) and (DcC) gives Ex. 2. Discuss the motion of a projectile having an initial velocity vo inclined an angle a with the horizontal, the resistance of the air being neglected. SolvJtUm. Assume the JTF plane as the plane of motion, OX as horizontal, and OY as vertical, and let the projectile be thrown from the origin. Suppose the projectile to be acted upon by gravity alone. Then the acceleration in the hori- zontal direction will be zero and in the vertical direction — p. Hence from (15), p. 106, Integrating, Vx = C\ and Vy = — gt-\-C%. But oo cosa = initial velocity in the horizontal direction, and Vosin a = initial velocity in the vertical direction. Hence C\ = Vo cos a and d = Vo ^^ a, giving (E) «x = »o cos a and Uy = — gft -f «o sin «• But from (10) and (11), p. 104, »x = — and »if = -^ ; therefore (E) gives dx J dy ^ , ' — =Vo cos a and — - = — fl« + to sin a, dt at or, dz= VoCOBadt and dy=z — gtdt + i^o Bin adt. 312 INTEGRAL CALCULUS Integrating, we get (F) X = »o cos tt • t + C» and y =:— \gfi + VoBma-t + Ci, To determine C3 and O4, we observe that when t = 0, jc = and y = 0. Substituting these values in (F) gives Cs = and C4 = 0. Hence (O) x = vo cos a • t, and (H) y = —\ffi^ + Vo^na't, Eliminating t between {O) and (H), we obtain (7) y = X tan a — gx^ 2ro*co82a which is the equation of the trqjectoryj and shows that the projectile will move in a parabola. EXAMPLES Find the function whose first derivative is x' 1. X — 3, knowing that the function equals when x = 2. Ans, -— — 3 x + 18. 2. 3 + X — 6 x^, knowing that the function equals — 20 when x = 6. , . . Ans, 804 + 3x + |--^- 2 8 3. (y* — &^), knowing that the function equals when y = 2. 4 2 4. sin a + cos a, knowing that the function equals 2 when a = -• Ans, sina — cosa + l. 6. 1 knowing that the function equals when t = 1. Ana. log (2 1 — t"). Find the equation of a curve such that the slope of the tangent at any point is 6. 3x-2. Ans. y = — -2x + C. 2 7. xy. Am. y = ce*. 8. x2 + 6x, the curve passing through the point (0, 3). ^ ^ ^iw. y = i. + -^ + 3. 9. -f the curve passmg through the point (0, 0). Ans, y> = 2px. y 10. — , the curve passing through the point (a, 0). Ans, 6«x* — a^ = a^. 11. m, the curve making ^n intercept b on the axis of y. ^n«. y = mx + b. CONSTANT OF INTEGRATION 818 Find the relation between z and y, knowing that 12. *? = ^. Ana. ^ = ^-^0. dx y 2 3 ' 13. xdy -f ydx = 0. Aiis, xy = C. 14. ^ = 1±-?, If y = when x = 0. Ana. x« + y« + 2jc - 2y = 0. dz 1 — y 16. (l-y)da;-f (l + x)dy = 0. Ana, logl±^ = C. 16. Find the equation of the cuire whose subnormal is constant and equal to 2 a. ^j. Ana. y^ = 4 ox + C, a parabola. JTiAl. From (4), p. 90, subnormal ^y—' dx 17. Find the curve whose subtangent is constant and equal to a [see (3), p. 90]. Ana. a logy = X + C 18. Find the curve whose subnormal equals the abscissa of the point of contact. Ana. y^ — x^ = 2 C, an equilateral hyperbola. 19. Find the curve whose normal is constant (= iZ), sssuming that y = jR when x = 0. Ana. x« + y« = iP, a circle. Biitt. From (6), p. 90, length of normal « y V 1 + (-^ j i or, dx = * ( JP - y^" ^ydy. 20. Find the curve whose subtangent equals three times the abscissa of the point of contact. Ana. x = cy^. 21. Show that the curve whose polar subtangent [see (7), p. 99] is constant is the reciprocal spiral. 22. Show that the curve whose polar subnormal [see (8), p. 99] is constant is the spiral of Archimedes. 23. Find the curve in which the polar subnormal is proportional to the length of the radius vector. Ana. p = ce°^. 24. Find the curve in which the polar subnormal is proportional to the sine of the vectorial angle. Ana. p = c—a cos 6. 26. Find the curve in which the polar subtangent Is proportional to the length of the radius vector. Ana. p = ce^. 26. Determine the curve in which the polar subtangent and the polar subnormal are in a constant ratio. Ana. p — ctf^. 27. Find the equation of the curve in which the angle between the radius vector and the tangent is one half the vectorial angle. Ana. p = c (1 — cos B). Assuming that v = Vo when £ = 0, find the relation between o and £, knowing that the acceleration is 28. Zero. Ana. o = 0o. 29. Constant = k. Ana. v — VQ + ld. 30. a + W. Ana. t = t>o-f crf4- — • 814 INTEGRAL CALCULUS Assuming that s = when t = 0, find the zelation between 8 and t, knowing that the velocity is 31. Constant (=«o). -^w- « = »ot. 32. m + iU. -Arw. « = m<-f — . 33. 3 + 2«-3t». ^»w. « = 3«-ft*-(\ 34. The velocity of a body starting from rest is 6 {* feet per second after t sec- onds, (a) How far will it be from the point of starting in 3 seconds ? (b) In what time will it pass over a distance of 360 feet meaflured from the starting point ? Ans, (a) 45 ft.; (b) 6 seconds. 36. A train starting from a station has after t hours a speed of ^ — 21 1^ + 30 1 miles per hour. Find (a) its distance from the station ; (b) during what interval the train was moving backwards ; (c) when the train repassed the station ; (d) the distance the train had traveled when it passed the station the last time. An8. (a) 1 1« - 7 1> + 40 <3 miles ; (b) from 6th to 16th hour ; (c) in 8 and 20 hours ; (d) 4658^ miles. 36. A body starts from O and in t seconds its velocity in the X direction is 12 e and in the Y direction 4<s - 0. Find (a) the distances traversed parallel to each axis; (b) the dis- tance traversed along the path ; (c) the equation of the path. An8, (a) x = 6f^ y = -<«-9t; (b), = lt« + 9t; (c)y = (^x-9)^|. 37. The equation giving the strength of the current i for the time t after the source of E.M.F. is removed, is (R and L being constants) Find i, assuming that I = current when t = 0. Ans. i = le l, 38. Find the current of discharge i from a condenser of capacity C in a circuit of resistance £, assuming the initial current to be Joi having given the relation di_ dt^ i " CjB' j_ C and R being constants. Ans. % = Jo^ci?. CHAPTER XXVI INTEGRATION OF RATIONAL FRACTIONS 188. Introduction. A rational fraction is a fraction the numerator and denominator of which are integral rational functions.* If the degree of the numerator is equal to or greater than that of the denominator, the fraction may be reduced to a mixed quantity by dividing the numerator by the denominator. For example, = ar-f a; — 3 + x^-\.2x-^l ar^-f2a:-fl The last term is a fraction reduced to its lowest terms, having the degree of the numerator less than that of the denominator. It readily appears that the other terms are at once integrable, and hence we need consider only the fraction. In order to integrate a differential expression involving such a fraction it is often necessary to resolve it into simpler partial frac- tions, i.e. to replace it by the algebraic sum of fractions of forms such that we can complete the integration. That this is always possible when the denominator can be broken up into its real prime factors will be shown in the next section.f 189. Partial fractions. Consider the rational fraction (A) ^^ reduced to its lowest terms, the degree of the numerator being less than that of the denominator. If a occurs a times as a root of the equation /(a;) = 0, we may write * That to, the Tariable is not aifected with fractional or negative exponents (see $25, p. 16). t Theoretically, the resolution of the denominator into real quadratic and linear factors is always possible when the ooeffl<;ients are real, that is, such a resolution exists. 816 J 816 INTEGRAL CALCULUS where <^ {x) is not divisible hj x — a. Therefore <^ (a) ¥= since a is not a root of <^ (2;) == 0. We may then write {A) in the form (5) ^<^> (a;— aY^{x) The equation F{x) ^ A F{x) A f{x) (a; — a)* (a;— a)«<^(a;) (a; — a)* is evidently true for any value of A, Combining the last two fractions, (0) J^(x)^ A F{x)-Ai>{x) ^ ' fix) {x-aY (x--a)-<t>{x) Now let us determine the value of A so that the numerator of the last fraction in ((7) shall be divisible by x — a. In that case x=za will be a root of F(x) — A<l> (x) = 0, and hence F{a)^A<l>{a)=0, or, A = ^. This value of A is finite since <f> (a) =^ 0. Having now determined the value of A so that x — a shall be a factor of F{x) — A^{x)^ we may write F{x) - A4>{x) = {x - a)F^{x), where the degree of the new function F^ {x) is less than the degree of (a:-a)«-*<^(a:). Hence from {C) f{x) {x-aY {x-aY-'4>(x) If a > 1, we proceed in the same manner with the second fraction on the right-hand side of (Z)), giving, say, where A. = ^^ . <^(a) . mXEGRATION OF RATIONAL FRACTIONS 817 Continuing this process, we get ^ ' /(x) {x-a)'^{x-a)'-'^ ^x-a^<l,{x) where G (x) is of lower degree than <t> (x), and <f> (x) does not contain the factor x — a. If now b OCCUI8 l3 times as a root of f{x) = 0, it will also occur 13 times as a root of ^(x)= 0, and we may break up GJx) <l>{x) into a sum of partial fractions in exactly the same way as we decomposed the original fraction. From the preceding discussion it follows that if a, J, <?,.., Z are roots of the equation /(a;) = 0, occurring respectively a, /8, 7, ..., X times, the given rational fraction may be broken up into a sum of rational fractions as follows : w S=7A-.+r-^+-+^ a-1 f{x) (x — a)' (z — a)*"' x—a + ^ + ?l + ... + ^£=1 {x-bf (x-bf-^ x-b "^ (a; _ Z)* "^ (X - i)*-' + ■ • • + ^37' Since every equation of degree n has n roots, it is evident that there will be as many partial fractions as there are units in the degree ot f(x). Instead of finding the constants A^A^^ • • •, J9, ^^ . . ., Z, Z^ --- by the method indicated above, it is more usual to clear ((7) of frac- tions. This makes F(z) identically equal to a polynomial in x of degree not greater than n — 1 [assuming n as the degree of F{x)] and therefore containing at most n terms. Since this identity holds true for all values of Xj we equate the constant terms and the coefficients of like powers of x on both sides. This gives n inde- pendent and consistent equations, linear in the constants required. Solving these n simultaneous equations we get the n constants 318 INTEGRAL CALCULUS 190. Imaginary roots. Equation ((7) on p. 317 holds true when some or all of the roots a, i, c, • - , 2 are complex (see § 11, p. 9). When the coefficients of F{x) and f{x) are real — and this is the only case we shall consider — we may avoid the complex numbers and put our results in real form as follows. Suppose, for instance, that the root b is complex. It may be written in the form j^^^;^,, where g and A are real numbers and i = V— 1. Then there must be a second complex root, say c, conjugate to (, namely, <? = ^ — hi. If h occurs fi times and c occurs 7 times, we see that fi must equal 7. From the manner in which J5, B^^ • • •, C, Cj, . . . were deter- mined it is evident that if J?= 6? -h Hi, B^ = G^ 4. HJ,, we shall have C^G-Hi, C^ = G^^H^i, Hence the sum '•» a a . . • x—h X— c may be expressed in the real form (B) t(£) , where the denominator is of degree 2/8 and the numerator of degree not greater than 2/8 — 1. Let '^i(:r) be the quotient and P^x-^-Q^ the remainder found in dividing '^{x) by [(x — ^)*-^ ^*]a Then ^ (^) ^ |-(^ _ gy ^ ^r| ^^ (^) ^ p^^ ^ ^^^ and {B) may be written in the form IC\ Vr(a:) ^ P^x + Q, ^,{x) ^ [(^-9f + hy [{x-9Y + h'Y^[{x-gY + h'Y-'' INTEGRATION OF RATIONAL FRACTIONS 819 Similarly the second fraction on the right-hand side of ((7) may be written in the form ilh V^i(^) ^ A^-hg, JTti^) Continuing this process, (A) gives finally ^ ' (x-bf'^ {x-bf-^'^""^ x-b + — ^—-\- ^ + --- + -^^ {x — cy {x — c)^^ x—c \{,x^gf^h^^^\{x-gfJr't^'Y-" ix^gf + h^' The coefficients P^ Q^, P„ Q^ . • ., P^, Q^ are calculated by the same method we employed to calculate the coefficients in the last section. We shall now proceed to illustrate what has been said by work- ing out numerous examples in detail. It is convenient to classify our problems imder the following four heads. Case I. When the roots of f{x) = are all real aod none repeated. The denominator may then be broken up into real linear factors, none of which are repeated. Case II. When the roots of f{x) = are all real but some repeated. Then the denominator may be broken up into real linear factors, some of which are repeated. Case III. When the equation /(a;) = has some imaginary roots, none of which are repeated. Then the denominator may be broken up into a product of real prime factors, there being a real quad- ratic factor (factor of the second degree) corresponding to each pair of conjugate imaginary roots. Case IV. When the equation f{x) = has some imaginary roots repeated. The corresponding quadratic factors are then repeated in the denominator. When we speak of factors of the denominator we shall mean only real prime factors, as these include all the types that can occur.* * tf the ooefflcient of the highest power of the rariahle in the denominator is different from nnity, we begin by dividing both numerator and denominator by this coefficient. 820 INTEGRAL CALCULUS 191. Case I. When the factors of the denominator are all of the first degree and none repeated. It is evident from ((y), p. 317, that to each non-repeated linear factor, such as 2; — a, there corresponds a partial fraction of the form A X— a Such a partial fraction may be integrated at once as follows : •/ x-^ a %/ X — a Ex. 1. Find C^l±3^. Solution, The factors of the denominator being x, x — 1, x + 2, we ajBSume* 2x + 3 A , B , C W Z7Z TTTII-r^ = T + x(x-l)(x-f2) X x-1 x + 2 where Aj B, C are constants to be determined. Clearing {A) of fractions, we get {£) ' 2x + 8=^(x-l)(x + 2)+B(x + 2)x + C{x-l)x = (A + B 4- C)x2 + (^ + 2 B - C)x - 2^. Since this equation is an identity, we equate the coeiBcients of the like powers of X in the two members according to the method of Undetermined Coefficients, and obtain the three simultaneous equations (C) Ja + 2B-C = 2, . l -2^ = 3. Solving equations (C), we get ^=~f, B=J, C = -f Substituting these values in (A), 2x + 3 __JL+ ^ ^ x(x-l)(x + 2) 2x 3(x~l) 6(x + 2) ^x(x-l)(x + 2) 2^ X 3^x-l 6^x4-2 = - } log*c + f log(x - 1) - i log(x + 2) + logc , c(x-l)« . x*(x + 2)* A shorter method of finding the values of ^, B, and C from (B) is the following: Let factor x = ; then 3 = ~ 2A^ or, ^ = — {. Let factor x — 1=0, orx = l; then 5 = 3 B, or, B = }. Let factor x + 2 = 0, orx = -2; then ~ 1 = 6 C, or, C = - J. *In the proc;e88 of decomposing the fractional part of the given differential neither the Integral sign nor dx enters. INTEGRATION OF RATIONAL FRACTIONS 821 ^ r (x-l)dx ^ J c(z -f4)t ' J x2 + 6x + 8 (x4.2)»* J x»-4x 3 2 * (x + 2)« . f x*dx X» _ 1, x-1 16, , . ov . r» 6. r <''-'')y''y iog('^-'')%c. /• (!P + pq)S U-p)(t + q) '• J t(rri,)>T^ - >°« i + ^• v^z*-5«« + 6 2V2 Z+V2 2V3 « + V3 192. Case II. When the factors of the denominator are aU of the first degree and some repeated. From (G)y p. 317, it is clear that to every Wrfold linear factor, such as (2; — a)", there correspond the n partial fractions (x — a)" (x — a)*^ x^a The last one is integrated as in Case I. The rest are all inte- grated by means of the power formula. Thus, J{x-ay J^ ^ (1 - w) (a; - a)-^ x« + l Ex. 1. Find f f ,,. dx SolutUm. Since x — 1 occurs three times as a factor, we assume x» + l ^ . B . C . D = r ■ 771 "T Z 77^ "" x(x-l)» X (x-l)» (x-l)« x-1 Clearing of fractions, x« + 1 = il (X - 1)» + Bx + Cx(x - 1) + I>x{x - 1)« =z{A +2))x« + (-8A + C-22))x2 + (3-4 + B-C + D)x--4. 822 INTEGRAL CALCULUS EqaatiDg the coefficients of like powers of x, we get the simultaneoas equations -4 + D = l, -3A + C-2D = 0, 3^ + B-C + l>=0, -^ = L Solving, -4 = - 1, B = 2, C = 1, D = 2, and x» + l 1. 2 , 1 . 2 x(a;-l)« X (x-l)» (x-l)« x-1 (X-l)2 ° X EXAMPLES 1. f -^ = _±_4.iog(x + l) + C. J (X + 2y(x + 1) x + 2*^ ' (X - 1)1 (» - 1)^ X - 1 * ^ ' ^ r(3xj,2)<ix^jlx4j_ _x^ J x(x + l)<» 2(x + l)a ^(x + l)a 5 r g'dx _ 6x + 12 /x4-4y _ * J (X + 2)2(x + 4)4 " x^-f 6x + 8"^ ^^\x + 2/ "^ ' 6. r ^^ = — ^ + log(y + 1) + C. •J(t^-2)2 4(^2-2) '^8V2''^t-V2 ' cui^ , . V . 2a4 8. I — -— = a log(« + a) H + C J (a-|-a)» *^ ^ « + o 2(a + a)2^ ^ \z + m (z + n)2/ ^'^ z + n 193. Case III. When the denominator contains factors of the second degree hut none repeated. From (E)^ p. 319, we know that to every non-repeated quadratic factor, such as a;*+jt>a;-f g, there corresponds a partial fraction of tiie form ^:r + ^ 3?J^px^q INTEGRATION OF RATIONAL FRACTIONS 323 This may be integrated as follows : / 4 (Ax + B) dx _ I V 2 2 y Ax + ^-^ + B"\dx Adding and subtracting — ^ in the numerator. I -f + B^dx — Ap\ /• dx _ A r (2x-\^p)dx / 2B '"2*/a?-f-jt>a? + 5' \ i)H'-^) [Completing tlie square in the denominator of the second integral.] ^ ^ /J. .v.2^— ^O ^ 2x-|-» ^ = TT ^og{ar-\-px + q)-i — ^ arc tan — Z= ^= -h C. Since :if -\r px -if q^^ has imaginary roots, we know from 3, p. 1, that4g-./>0- X Ex. 1. Find f-r^ Solution. Aflsame — = — | — x(x2 4.4) X x2 + 4 Clearing of fractions, 4 =-4 (x« 4- 4) + x(Bx+ C)= (-4 + B)x2 + Cx + 4A. Equating the coefficients of like powers of x, we get -4 + 5 = 0, C = 0, 4^=4. 4 1 X This gives A = 1, B = - 1, C = 0, so that = — * » » » x(x2 + 4) X x2 + 4 /4 dx __ rdx ^ r zdx x(xa + 4)"J 'x"'~Jx« + 4 = logx — - log(x2 + 4) + log c = log ^ . il?w. 2 VxM-4 EXAMPLES , r «lx 1 , x« + 4 2 ^ X . ,7 1. I = — log 4- - arc tan - + C. J (X + 1) (X» + 4) 10 * (X + 1)8 6 2 ^ r (2x«-3x-8)dx , (x«-2x + 6)* 1 ^ x-1,^ 2. I -^ = log ^ + - arc tan \- C. J (x-l)(xa-2x + 6) ^ x-1 2 2 „ r x«dx 1 , 1 + X 1 ^ . ^ 3. I = - log arc tan X + C J 1-x* 4 ^1-x 2 824 INTEGRAL CALCULUS . r dx 1, aj* 1 . _ 4. I — = T log = arc tan x + C. 5- I -4 — TT-i — :: = log , + - arc tan arc tan — — + C. r (6x' — l)dx , X*— 2x4-6 5 x—l 2 x ^ 6. I — ^ = log — h -arc tan -arctan^-+C J (xa+3)(x2-2x+6) *x2 + 32 2 Vs Vs ^ f dx 1. (x + 1)" 1 , 2x-l , ^ 7. I — = - log -^ h — - arc tan — - — \- C. Jx«-4-l 6 *x8-x+l Vi V3 8. I — ^ — = log + - arc tan - + C. Jx« + x2 + 4x4-4 *(x + l)^ 2 2 9. I = - log I 1 + -z- arc tan — - + C 10. I = — - log — ;= h V 2 arc tan h C ,, r dy 1, i^ + y + 1 . 1 ^ 2y4-l . ^ 11. I __*-- = - log -| — ^ r + -^arctan ^ + C. 194. Case IV. TF%6n the denominator contains factors of the second degree some of which are repeated. To every w-fold quadratic factor, such as (a?-fj9a;+g)", there correspond the n partial fractions /J\ ^^4-^ . Cx + I) Lx^M To derive a formula for integrating the first one we proceecl as follows : /; Ax + f-f + B^dx {a?+px + q)' "" ~J {3»+px + qy ^ + S . . _ V " 2 2^ J I Adding and subtracting -^ in the numerator. I Ax + ^^dx C(-^ + B\dx (7?^px+qY J {3? + pxJrqY d C,-^ ^ „. ^ .^-./o . ^ „x J. ^ /2B-Ap\ C dx J(.. +;,x + ,r(2 - +l>)ix + (i«^)/^^ ^^^ ^ ^^. INTEGRATION OF RATIONAL FRACTIONS 826 The first one of these may be integrated by [4] ; hence (^ /i Ax-^-B dz = {J? +px + qY 2 (1 - n) (a? J^px + qf'^ dx + (^^)/; {a?^px^qy hiL. Let us now dijBEerentiate the function -,, {^ ■¥ px -\' qy-"^ Thus, - + | \ 1 2(n-D(.+f S dx \{2^'\'px -f g)*"7 {^ -i-px + q)*~^ {3? +px + qy (0) . d ^2 {^'¥pX'\'qT-\ {3?+px + q)'-^ (sd'+px + q)' / fsiaoe 3fl+px+q~(x+^+(q-^, «nd («+^ -(«'+J»*+«)-(«-^-l Integrating both sides of (<7), P « + *- 2 .„„./» da; + 2(„-l)(,-f)/- (fi; or, solving for the last integral, J{^+px^qr 2in-^l)(q^fji:^^px-^qr^ -f or. 2n-3 r dx 326 INTEGRAL CALCULUS Substituting this result in the second member of (5), we get* {Ax + B)dic _ A(p*^4:q) + {2B^Ap)(23D+p) (ic*+px + q)^ 2(n — l)(4g —!>■){»■+ pa? + gr-* die mf; (2B - Ap) (2n - 3) r "*■ (n-l)(4g-i>*) J {X* + px + q)*^-^ It is seen that our integral has been made to depend on the integration of a rational fraction of the same type in which, how- ever, the quadratic factor occurs only n — 1 times. By applying the formula (E) n — 1 times successively it is evident that our integral may be made ultimately to depend on dx f-. 0^ -\-px-\-q and this may be integrated by completing the square, as shown on p. 802. In the same manner all but the last fraction of {A) may be integrated. But this last fraction, namely,* Lx-\'M —5 1 or '{'px-\'q may be integrated by the method already given under the previous case, p. 823. Ex. 1. Find / (a;« + a;« + 2)(iaj {x2 + 2)2 Sohdion, Since 2^ + 2 occurs twice as a factx)r, we assume g» -i- x^ + 2 _ Az-\- B Cx-\- D (x2 + 2)2 " (x2 + 2)2 "^ xs + 2 * Clearing of fractions, we get x» + x2 + 2 = ^x + B + (Cx + D) (x2 + 2) = Cx8 + Dx2 + (^ + 2 C)x + B + 22X Equating the coefficients of like powers of x, C = l, D = l, ^ + 2C = 0, B+2D = 2. This gives ^ = - 2, B = 0, C = 1, D = 1. „ x8 + a;2 + 2 2x x + 1 Hence — ^^— = 1 — — — , and (x3 + 2)2 (x2 + 2)2^x2 + 2 / (xg + x2-i-2)dz _ _ r 2zdx r xdx r dz {X2 + 2)2 "J (X2 + 2)2 "^ J X2 + 2 "^ J X2^ 2 1 X 1 + -— arctan— - + -log(x2 + 2) + C «^ + 2 ^^ \/2 2 • 4g-/>«>0, since x*-^px + q=0 has imaginary roots. INTEGRATION OF RATIONAL FRACTIONS 827 Ex. 2. Find fg^ + ^ + Sdx. Solution. Since x^ _|. i occurs twice as a factor, we assume 2x* + x + S _ AX'{-B Cx + D {x«+l)a ~(x2 + i)a'*" a^ + i * Clearing of fractions, 2x« + X + 3 = -4x + 5+ (Cx + D)(x2 + 1). Equating the coefficients of like powers of x and solving, we get ^=-1, B = 3, C = 2, D=0. Hence r2x»+^+8 ^ f-^ + S /• 2xdx •^ (x2 + 1)2 J (x» + 1)« J xa + 1 = log(xN-l)+/^,dx. Now apply formula (J?), p. 326, to the remaining integral. Here -4 = - 1, 5 = 3, i) = 0, g = 1, n = 2. Substituting, we get /•-X4-3, l + SxSfdx 1 + 3x3 I — -ox = + - I = + - arc tan x. J (x2 + 1)4 2(x2 + 1) ^ 2 J x2 + 1 2(x2 + 1) 2 Therefore f2x8 + x + 3 I , o . ^x l + 3x 3 I — rs — TTTT- = log(xa + 1) + — -T + - arc tanx + C. J (x2+l)a '^^ ^ ' 2(x2 + l) 2 1. r?l±£ — }_^ _ _? — ?_ _^ iQg /^.a ^ 2)* — arc tan — + C. J (x2 + 2)a 4(x2 + 2)^ ^^ ^ ' 4v^ V^ 2. r — ^^^ =i,og^L±i+_^ij_+c. J (1 + x){l + x2)a 4 ^ (X + 1)2 2{x» + 1) rx7 + x» + x« + x ^ 6 _10_ l»j^^ 2 2) - 91og(x2 + 3) + C. J (xH2)2{xa+3)2 2(xa + 2) x3 + 3 2 *^ ^' 6V t/t . f (4x2-8x)dx 3x2-x . (x-l)a . ^ .^ 4. I — ^ = hlog^^ ^ + arctanx + C. J (X- l)a(x2 + 1)2 (X - 1) {x2 + 1) ^ x2 + 1 - r (Sx-\-2)dx 13X-24 .26 ^ 2x-3 . _ 6. I — ^^ = J- arc tan \- C. J {x2 - 3x + 3)2 8(x2 - 3x + 3) 3 V3 V3 Since a rational function may always be reduced to the quotient of two integral rational functions (§ 26, p. 16), i.e. to a rational fraction, it follows from the preceding sections in this chapter that any rational function whose denominator can be broken up 828 INTEGRAL CALCULUS into real quadratic and linear factors may be expressed as the algebraic sum of integral rational functions and partial fractions. The terms of this sum have forms all of which we have shown how to integrate. Hence the Theorem. The integral of every rational function whoBe denomi- nator can he broken up into real quadratic and linear factor% may he founds and is expressible in terms of algebraic, logarithmic, and inverse-trigonometric functions; that is^ in terms of the elementary functions. CHAPTER XXVII INTEGRATION BY SUBSTITUTION OF A NEW VARIABLE. RATIONALIZATION 195. Introduction. In the last chapter it was shown that all rational functions whose denominators can be broken up into real quadratic and linear factors may be integrated. Of algebraic functions which are not rational^ that is, such as contain radicals, only a small number, relatively speaking, can be integrated in terms of elementary functions. By substituting a new variable, however, these functions can in some cases be transformed into equivalent functions that are either in the list of standard forms (pp. 292, 293) or else are rational. The method of integrating a function that is not rational by substituting for the old variable such a function of a new variable that the result is a rational function is sometimes called integration by rationalization. This is a very important arti- fice in integration and we will now take up some of the more important cases coming under this head. 196. Differentials containing fractional powers of x only. Siich an expression can be transformed into a rational form by means of the stibstitution a: = e", where n is the least common denominator of the fractional exponents ofx. For x, dx^ and each radical can then be expressed rationally in terms of .. —dx. Solution. Since 12 is the L.C.M. of the denominators of the fractional expo- nents, we assume x — z^ Here dx = 12 z^kte, x' = z^, x* = 2», x* = sfi. .-. r ^ ~^ dz = r ^~^' 12 2"dz = 12 Hzw - z8) dz = f z" - I z« + C = f X* - 1 X* + C. fSabetitating back the value of z in terms of x, namely, z=x'^.l 329 830 INTEGRAL CALCULUS The general form of the irrational expression here treated is then I where R denotes a rational function of o^. 197. Differentials containing fractional powers of a + to only. Such an expression can be tran^ormed into a rational form by means of the substitution a-{-bx = z\ where n is the least common denominator of the fractional' exponents of the eapression a -\- bx. For Xy dxj and each radical can then be expressed rationally in terms of .. Ex. 1. Find ^ ^ r ax •^(H-x)* + (l + x)*' Solution. Assume 1 + x = 2^ ; then dx = 2 zdz, (1 + x)' = ««, and (1 + «)* = «. dx r 2zdz _ r dz (1 + a;)i -f (1 + X)* ""'^ «• + « ~ J z^-hl = 2 arc tan 2 + C = 2 arc tan (1 + x)* + C, when we substitute back the value of z in terms of x. The general integral treated here has then the form I -B [a:, (a -h bxy] dx^ where B denotes a rational function. '}dz 4 ;*-x* , 1/2 . 6 6x^ ;* + 1 ^ 6 12 3. r^Jl-dx = --- + — + 21ogx-241og(xA + i) + C. •^ X* + X* X* X" /elf ft f^ 1 — = -X* + 2 log "" + 4 arc tan x* + C. X" — X* «* X* + 1 SVxdx ,«rx' x' 4x* /avxox ,„rx* x« 4x' 11 1 T INTEGRATION BY RATIONALIZATION 831 •^aj{x + l)* (x + l)*H-l "•^(o + te)* ftaVa + to ' '•^(4x + l)* 12(4x + l)* 0. Cy Vo H- y dy = A (4 y - 8a) (a + y)* + C. 1. r ,^jti'^^ dx = x + l + 4Vx + l + 41og(Vx + l^l) + C. •^ Vx + 1 - 1 2. f ^= = ?{x + 1)« - 3(x + 1)* + 31og(l +^^iTl) + C. •^ 1+Vx + l 2 . f ^ 1 = 8 {(X + 1)* + 2 (x + 1)» + 2 log [(X + 1)* - 1]} + C. •^ (X + 1)' - (X + 1)* L98. Differentials containing no radical except Va + bx + acK* Such an eapression can be transformed into a rational form by means of the substitution Va -\'bx-\-3[? = 2 — a:. For, squaring and solving for x, ^ = ^; thenefe=2(zl±^^ and Va4-6x + a:'(=g~a?)= , J" — ^ b-\'2z Hence x^ dx^ and Va -\-bx-\-:3^ are rational when expressed in terms of z. -== V 1 + X + X* Solution. Assume Vl + x + x* = « — x. Squaring and solving for x, 2z + l {22 + 1)-^ and vT+x + x2(==2--x)== V^— ^. ^ ' 2« + l • If the nullcal is of the form Vn +px + qx*^ (7>0, it may be written Vgx/'* + ^x + x«, and therefore oomes under the above head, where a = - 1 6 = - • 9 9 332 INTEGRAL CALCULUS jf::!>; =/-f^=iogc(2,.n)c] 2« + l ^ = log[(2x + 1 + 2 Vi + a; + x«)c], when we substitute back the value of z in terms of x. 199. Differentials containing no radical except Va + 6x — «■.* aSWcA aw expression can be transformed into a rational form by means of the substitution •\/aJ^bx-a? [= V(a: - a) (/S - a:)] = (x - a) 2, wAere a; — a an<i /8 — a: are real^ factors of a-\-bx — a^. For, if -Va-^-bx — a? = V(a: — a) (/8 — a;) = (a? — a)^, by squaring, canceling out {x — a), and solving for a;, we get a: = ^;theni. = ^(;^-y , and VaHri^^^[= (a: — a)2j] = ^ ,"" V^ * Hence «, rfa;, and Vo-f-fti^-^ are rational when expressed in terms of 2. dx /ax V2 + X - x* ^Solution. Since 2 4- « - a:* = (x + 1) (2 — x), we assume V(x + 1) (2 - X) = (X + 1)2. 2 - 2« Squaring and solving for x, x = — — - . 2 "T" A -62d2 __../:;—-: — -Sr . . .. , 3« — Q ZtiZ / Hence dx = — - — — » and v2 + x - x« [= (x + 1)2] = (22 + 1)3 '■''"' 2« + l /dx ^ r dz -;== = - 2 I -— - =- 2 arc tan2 + C V2 + X - xa ^ 2« + 1 = — 2 arc tan \/ + C, when we substitute back the value of z in terms of x. /2~x • If the radical is of the form Vn+px - qjc», g >0, it may be written V? \/- + - x - «•, and therefore oomm under the above head, where a=-ib=-- g 9_ t If the factors ota-i-bx-x* are imaginary, Va + 6x-ar* is imaginary for all ralaes of x. For, if one of the factors is x - m + in, the other must be - (x - m - in)^ and therefore 6 + ox -x* = - (x-m-¥- in) (x-m-in) — - [(x—m)* + n*], which is negatiye for all values of x. We shall consider only those cases where the factors are reaL INTEGRATION BY RATIONALIZATION 833 EXAMPLES Vxa - X + 2 +a:-V2 , /• dx 1 , Vxa - X-I-2+X-V2 ^ _ 1. ) — ; = "-F tog -7: 7: + C. •^ X V x* - X + 2 V 2 V x2 - X + 2 + X + V 2» dx 2. f — ^ = 2arc tau(x + Vx2 + 2x-l) + C, ^ xVxaH-2x-l 3. r__^^^ = _2arctanfLz^y + C. •^ V2 - X - x2 \x + 2/ dx 4. f ^ = log(2 Vx5« - X - 1 + 2x - 1) + C. •^ Vx2 - X - 1 * r <^ « /l — as ^ 6. I -—== = - 2arc tan -v/ ^ + C ^ V4x-3-xa \x-3 g r xdx ^ 8-K6X ^^ * -^ (2 + 3x-2a^)* 26V2 + 3x-2x» 7. r ^<^ = V6 log(6x - 1 + V6 V6xa-2x + 7) + C. ^ V6xa-2x+7 8. f ^ — = -i,log(6x - 1 + 2 V3 V8x2-xHH) + C. •^ V3x»-x + l V3 9. f-— i^^=2%^arc8infi^) + C. •^ V4H-3x-2x^ . ^ V4I ^ 10. r_^=, = log(l + x + ViM^) + C. 11. r<lfL±^)^ = log(x + l+V2^T7a) i -fC. •^ x^ X + V2x + xa The general integral treated in the last two sections has then where R denotes a rational function. Combining the results of this chapter with the general theorem on page 328, we can then state the following Theorem. Every rational function of x and the square root of a polynomial of degree not higher than the second can be integrated and the result expressed in terms of the elementary functions.* * As before, however, it is assumed that in each case the denominator of the rational function can be broken up into real quadratic and linear factors. 334 INTEGRAL CALCULUS 200. Binomial differentials. A differential of the form where a and b are %ny constants and the exponents m, n, je> are rational numbers, is called a binomial differential. Let a: = z*; then dx = a«*"*(fe, and or (a 4- bif'ydx = a«-^+— ^(a + bsT^ydz. If an integer a be chosen such that ma and na are also integers,* we see that the given differential is equivalent to another of £he same form where m and n have been replaced by integers. Also, or (a -h b3fydx = oT'^'^iax-'' + bydz transforms the given differential into another of the same form where the exponent n oix has been replaced by — n. Therefore, no matter what the algebraic sign of n may be, in one of the two differentials the exponent of x inside the parenthesis will surely be positive. When |> is an integer the binomial may be expanded and the differential integrated termwise. In what follows p is regarded as a fraction; hence we replace it by -, where r and % are integers.! We may then make the following statement : Every binomial differential may be reduced to the form r of (a -{- bif)' dxj where w, w, r, 8 are integers and n is positive, 201. Conditions of integrability of the binomial differential r (A) «"* {a + 6x~)* dx. Case I. Assume a-^-baf = 2^, 1 r Then (a + baf)' = 2, and (a -f bsf)' = s^; I m also x=( T" j , and af* = ( ~ j ; hence dx = -r-^'^ ( 7 1 ^^• 071 \ b J * It is always possible to choose a so that ma and na are integers, for we can take « as the L.C.M. of the denominators of m and n. t The case where p is an integer is not excluded, but appears as a special case, yis., r»p, • « t. I a INTEGRATION BY RATIONALIZATION 835 Substituting in (A)^ we get The second member of this expression is rational when Tn + l n is an integer or zero. Case IT. Assume a-^-baf" = ^'af. Then af = r » and a -h 62f = ^^af = r r r Hence (a + bo*)' = a'(z^- b) V; 1 1 M m also a; = a»(2'-6)"% a:" = a"(2-- J)"»; « 1 -1-1 and (fo = --a"2;-^(2;'-J) " rfa?. n Substituting in (A)^ we get 8 —r- + z + I . *• /ne-\- 1 . r a:-(a + 6af)'rfa; = --a " •(z'-J) v « . ^z--*-'-^dz. n The second member of this expression is rational when — — — \- - . . n 8 IS an mteger or zero. Hence the binomial differential a:^ (a -f b^ydx can be integrated by rationalization in the following ca8e8:* Case I. When — ^^^— = an integer or zero^ by a88uming n a -{- baf* =^ z^. m -I- 1 ?* Case II. When — — — \.- = an integer or zero^ by a88uming n 8 a -f bsf = z*a:*. * Annmlng as before thAt the denominator of the resulting rational function can be broken up into real quadratic and linear factors. L 886 INTEGRAL CALCULUS EXAMPLES •^ (a + tea)* ^ ^ Vo + tea So2u^ion. m = 8, n = 2, r= — 8, 8 = 2; and here = 2, an integer. Hence this comes under Case I and we assume *'' — - — ) , dx = -; -1 and (a + te*)" = «•. / x^dz _ r/ z^ — a \* «2z 1 (a + te«)*~*^^ * ^ 6*(za-a)* «• 1 2a + 6x2 ^ Va + te» + C. " ^x* dx ^ (2xa-l)(l+x«)* VT+^'" 3x» + C. Solution. m = -4, n = 2, - = — j and here 1- - = — 2, an integer. 8 2 n s Hence this comes under Case II and we assume l + x« = «2x2, « = (L±^. X ' whence «■ = -:; — r» l + x« = -- » Vl + x^ = 2a_i ^ 2a_i ^ («'-l)* 1 1 , afa also X = , X* = — ; and dx = — zdz r dx I (2*"1)' /• (Z«-l)« (,S_1)1 _ z» , ^_ (2x'-l)(l + x«)* , „ •^ 16 4. C^L- = ..^^=^ + c. 6. r-^ = (i + «.)i<^ + c. •^ (1 + x2)' Vl+x2 -^ Vl+xa 3 ^ r dx 1, ex 6. j — -p=:^=i = -l0g- •^ X V a2 - x« « Va2 - x^ + a 7. r__^^ = -a(l+xT*(2x + l) + C. INTEGRATION BY RATIONALIZATION, 887 9. rt»(l + 2«2)^(tt = (l + 2«»)«^^- — ?:+C. 10. ru(l + w)*du= A(H-u)*(5u-2) + C. a^da a* 11. f-^ + C. 6a2)* 3a(a + 6o2)* 12. /(^ (1 + <^' (W = 5»j (1 + <^) V - i (1 + ^)» + A (1 + ^* + C. dx _ 3g» + 2a 13. r ^ , = "-^^^^^ -fC. 202. Transformation of trigonometric differentials. From Trigonometry (A) sin a: = 2 sin -cos-, 87, p. 2 (B) cos X = cos* - — sin* - . 87, p. 2 a? jy , , X 1 1 2 But sm - = 2 x •^^2 V^^'l-^' >/l+*^ X 1 and cos - = - 2 If we now assume -w sec^ 'l+tan*| tan ^ = 2? or, x = 2 arc tan 2, X Z X A. we get sin - = , , cos ^ = , Substituting in (A) and (5), 2^ 1-2* sma: = r ;, cosa; = 3 ;• 2d2 Also by differentiating x=2 arc tan z we have dfx = 3 ^ 388 ^ INTEGRAL CALCULUS Since sin Xj cos s^ and dx are here expressed rationally in terms of 2, it follows that A trigonometric differential involving sinx and cosx rationally only can be trantformed by means of the mbstitution tan- = j5, orj what is the same thing^ by the svhstitutions 2z 1-z^ J 2dz sm2:=:3 :;> cosa: = r r, aa; = 1+2* 1+2" 1+2" into another differential expression which is rationed in z. It is evident that if a trigonometric differential involves tan z, cot Xj sec x^ CSC x rationally only, it will be included in the above theorem, since these four functions can be expressed rationally in terms of sin Xj or cos x^ or both. It follows therefore that any rational trigonometric differential can be integrated.* EXAMPLES I. I ^ ^ ^ =-tan«- + tan-4--logtan^ + C. J sin X (1 + C08 sc) 4 2 22^ 2 (1 + COB x) SoliUum, Since this differential is rational in sin x and cos x, we make the ahore substitutions at once, giving / (l + 8inx)dx _ / \ l + z'/14-a sinx(l + co8x) " I 22 / 1 -gg •^ l + zA "'"l + ga ) 1 *X X 1 . / x\ ^ tan? -2 2. r_^=!iog_J— +0. •/4 — 6smx 3 1^. X .1 2tan--l 2 • See footnote, p. 330. INTEGRATION BY RATIONALIZATION 889 3. f = -arc tan (^2 tan -Wc. J 6-3C08X 2 \ 2/ . /• dx 1 ^ /6tanx + 4\ , ^ 4. I -——-— = - arc tan ( ) + C. dx 1 5 C = - arc tan(3 tan x) + C. J6-4co62x 8 ^ ' tan- + 2 6. (- — = 7log +C. ^3 + 6coflX 4 \ X „ tan — 2 2 /Binxdx 2 , _ * /* asx , ^ = _^_ + 2 arc tan ( tan - J + C l+sinx . . , X \ 2/ 1 + tan- ^ /• C0BX(2x ^ / x\ X « a I = 2arctan(tan-)-tan- + C. Jl + cosx V 2/ 2 9. Derive by the method of this article formulas [16] and [17], p. 208. 203. Miscellaneous substitutions. So far the substitutions con- sidered have rationalized the given differential expression. In a great number of cases, however, integrations may be effected by means of substitutions which do not rationalize the given differen- tial, but no general rule can be given, and the experience gained in working out a large number of problems must be our guide. A very useful substitution is 1 , dz Z 2* called the reciprocal substitution. Let us use this substitution in the next example. Vaa-x» -ox. Ex. 1. Find /— ^ X* I dz Solution. Making the substitution x = -f dx = — -,weget z z* J X* */ ^ ' SaJ 3aax» 840 INTEGRAL CALCULUS 1. f — — = — log +C. AaBamex' = £. Jx(a» + x») 3a« *o« + x« AflBume X — 2 = z. + log(x + 1) + C. Assume x + 1 = £• 6{x + l)8 — — — = ^ + C A88nmex=:- _ /• dx 1, ex 6. I , = -log •^ X Va« + x* <» a + V o2 + x« Assumo X = — • 6. f-^^ = |(x«-8)(x« + l)* + C. Assume x* + 1 = «. •^ (x^ + 1)* 8 /(2x ex 1 — - = log - » Assume x = -' X Vl + X + x2 2 + x + 2vl+x + x^. « 8. /^-ti?i^dx = ?(l + logx)* + C. Assume 1 + log x = «. X u = --(3c* - 4) (c + 1)« + C. Assume c* + 1 = «. (e* + l)* 21 1^' f-i ;r- = :;^ ^ + -log(e«-2) + C. Assumee« = «. J ^x_2e* 2e* 4 4 ' /x'^dx 1 X Vl jC* , = - arc sin X h C Assume x = cos z. Vl-x« 2 2 va* — x"dx = — arc sin - -(- - va'-' _ x^ + C. Assume x = a sin t. 2 a 2 CHAPTER XXVIII INTEGRATION B7 PARTS. REDUCTION FORMULAS 2S01 Formula for integration by parts. If u and v are functions of a single independent variable, we have from the formula for the differentiation of a product, V, p. 144, d{uv) = lulv + vduj or, transposing, udv = d (uv) — vdu. Integrating this, we get the inverse formula, {A) I udv = ut; — I vdUf called the formula for integration by parts. This formula makes the integpration of udvj which we may not be able to integrate directly, depend on the integration of dv and vdu^ which may be in such form as to be readily integrable. This method of integror Hon hy parts is one of the most useful in the Integral Calculus. To apply this formula in any given case the given differential must be separated into two factors, namely, u and dv. No general directions can be given for choosing these factors, except that (a) dx is always a part of dv ; and (6) it must he possible to integrate dv. The following examples will show in detail how the formula is applied. Ex. 1. Find jxcoBxdx. Sclviion. Let u = x and do = cos idx ; then du = dx and v = fcos xdx = sin x. Sal)Btitating in {A), u dv u V V du z C08 xdx = X sin x — I sin x dx = X sin X + cos X + C. 341 332 INTEGRAL CALCULUS / 2(z« + z + l)dg = log[(2x + 1 + 2 Vl + x + x»)c], when we substitute back the value of z in terms of x. 199. Differentials containing no radical except Va + bx — as*.* aS'uc^A aw expreBsion can he transformed into a rational form by means of the suistittUion Va-l-62: — a:*[=V(ic -~a){^ — x)] = (x — a)«, wAer« a; — a a?ui )8 — a; are real ^ factors of a + hx — of. For, if Va -f Jar — ar* = V(a; -^a){fi^ x) = (2; — a)2, by squaring, canceling out (a; — a), and solving for x, we get and Va -f 6a: - a:^r= (a? - a)2] =i^fl^. Hence x^ dx^ and Va -{-bx^a? are rational when expressed in terms of 2. /dx -7 V 2 + X - x^ (SoZti^ion. Since 2 + x — x^ = (x + 1) (2 — x), we assume V (X + 1) (2 - X) = (X + 1)2. 2-«2 Squaring and solving for x, x = z^+l Hence dx = /^^f , and V2 + x-x«[= (x + 1)«] = ^' . = - 2 I -«—T =-2arctaji2 + C V2 + X - xa -^ 2^ + 1 , 12 —X = — 2 arc tan \ H C, \x + 1 when we substitute back the value of z in terois of x. • If the radical is of the form Vn+px - qx^, 9 >0, it may be written y/q \/^ + - x - «•, and therefore oomea under the above head, where a= - f 6= - • q q__ t If the factors of a + 6ar - ar* are Imaginary, Va+6a:-jr» is imaginary for all values of x. For, if one of the factors is x - m + in, the other roust be - (x - m - xn)y and therefore 6 + ax-x* = -(x-m + \n^ (x-m- in) = - [(x - m)* + n*], which is negative for all values of x. We shall consider only those cases where the factors are real. INTEGRATION BY RATIONALIZATION 333 1. I —. = = -7=^<^g" 7 F+^- •^ X Vx* - X + 2 v2 V x2 _x-f2 + x + v2» 2. f — ■ = 2arc tau(x 4- Vx^ 4- 2 x - 1) 4- Cf. •^ xVx2 + 2x-l 3. f ^ =-2arctan(i— ^) + C. •^ V2 - X - x^ \x + 2/ 4. f , ^ - = log{2 Vx'* - X - 1 4- 2x - 1) + a •^ Vx« - X - 1 5. r , ^ — =~-2arctan^/l^-l-rf. •^ V4x~3-xa \x-3 /• xdx 8 4- 6x ^ ^ (2 + 3x-2x2)i 26V2 + 3x-2x« 7. f , ^^ =V51og(fix-l^V5Vfix2-2x4.7)^rf. ^ V6xa-2x+7 •^ V3x«-x + l v3 9. f A^__=2V2arcBin(i^Uc. •^ V4 + 3x-2x2 . ^ Vil ^^ 10. f -^^=^ = logf^ + X + Vx» + x) + C. •^ Vx« + x ^2 / 11. r(lfL±^)i^ = iog(x + i+V27T7«) i + c. -^ xa x + V2x-f xa TThe general integral treated in the last two sections has then the form ^ ^^ -sla^hx-^c^) dx, where R denotes a rational function. Combining the results of this chapter with the general theorem on page 328, we can then state the following Theorem. Every rational function of x and the square root of a polynomial of degree not higher than the second can be integrated and the result espressed in terms of the elementary functions,* • Ag before, however, it is asanmed that in each caae the denominator of the rational function can be broken up into real quadratic and linear factors. 344 INTEGRAL CALCULUS 13. fx.e«d, = !^Cx.-?^ + »f-A) + c. J a \ a a> a'/ 14. r 0> sin 0d0 =:2co8 + 20Bin0-0Sco8 + C. 16. r(logx)a(to = x[loga«-21ogx + 2]4-C. /ft* a tan* ada = a tan o — — + log cos ft + C "■/^^■-.-fT'°" -■"«'- ■> + <'• dx Bint. Let u=« log x and d» =• 1 etc. /x' X* + 2 / x«arc8inx(ix = --arc8inx4- — ^— vl-x«+ C. 3 19. r8ec9 0logtan9d« = tantf(logtan^- 1) + C. /dx log (log x) — = log X • log (log x) — log X + C. X 21. rM+il^ = 2 v5rTi[iog(x+i)- 2] + c. •^ Vx + 1 22. fx^{a - x«)*dx = ~ J x2(a - x«)« - ,J^ (a - x2)i + C. lRn<. Let u => a;> and dv » (a - a;*)' xdXt eto. •^ x' Sx^L ^ ^J 205. Reduction formulas for binomial differentials. It was shown in § 200, p. 334, that any binomial differential may be reduced to where p is & rational number, m and n are integers, and n is posi- tive. Also in § 201, p. 334, we learned how to integrate such a differential expression, in certain cases. In general we can integrate such an expression by parts, using (-4), p. 341, if it can be integrated at all. To apply the method of integration by parts to every example, however, is rather a long and tedious process. When the binomial differential cannot be integrated readily by any of the methods shown so far, it is cus- tomary to .employ reduction formulas deduced by the method of integration by parts. By means of these reduction formulas the given differential is expressed as the sum of two terms, one of REDUCTION FORMULAS 345 which is not affected by the sign of integration, and the other is an integral of the same form as the original expression, but one which is easier to integrate. The following are the four principal reduction formulas. [A] Cx^ {a + hx*^)P dx [E] Cx"^ (a + bx^)Pdx = ^ — ' — f- ^- I a^ (a + hx^\P'-^dx% np + t» + 1 np + «» + 1 ^ [C] Cx^(a + bx^)Pdx (w + l)a (w + l)a J [I>] fa?"* (a + hx'^)P dx n(jp + l)a n(p + l)a J While it is not desirable for the student to memorize these formulas, he should know what each one will do, namely: Formula [A] diminishes mhy n. Formula [J5] diminishes p hy 1, Formula [C] increases m hy n. Formula [I>] increases p by 1. I. To derive formula [A], The formula for integration by parts is (A) j udv = uv-' i vdu. (-4), p. 341 We may apply this formula in the integration of CQir{a-\-h7rYdx by placing u = a:^""^^* and dt; = (a 4- bafyaf^dx; then (iw = (7n-n + l)rr- ia: and v = ^?-t_^!)^. nb{p-^l) * In order to integrate dv by [4] it is necessary that x outside the parenthesis shall have the exponent n -1. Subtracting n - 1 from m leaves m-n + l for the exponent of x in u. 846 INTEGRAL CALCULUS Substituting in (A)^ (B) Car {a -{- baf'y dx wi(j? + l) nb{p + l)J ^ ^ But (V -" (a -f 6x-)'» +^ (ia; =f^~* (« + ia:")" (« + ia:") dx = a Ta:*— (a -f 6af')''cfo + h]Qr{a + harydx. Substituting this in (5), we get Transposing the last term to the first member, combining, and solving for |V'(a 4- haf^ydx^ we obtain [A\ fx^ {a + bx^)P dx b(np + m + l) b{np + m + l)J k t ^^ } It is seen by formula [A] that the integration of af"(a 4- la^ydx is made to depend upon the integration of another differential of the same form in which m is replaced by m — n. By repeated appli- cations of formula [-4], m may be diminished by any multiple of n. When rip 4- w -h 1 = 0, formula [-4] evidently fails (the denomi- nator vanishing). But in that case m4-l ^ n hence we can apply the method of § 201, p. 336, and the formula is not needed. IL To derive formula [B], Separating the factors, we may write ((7) (3r{a 4- ha^ydx =f^{a 4- baf^y-^ (a 4. baf) dx = a Car (a 4. bafy^dx 4. bCar+''{a 4- 6a:")''-^ia;. REDUCTION FORMULAS 847 Now let us apply formula [A] to the last term of (C) by substi- tuting in the formula m + n for m and j? — 1 for p. This gives Substituting this in ( (7), and combining like terms, we get [B] Cx^ (a + 6a5*») p da> = ^ — • ^—4- ^- \ x^(a 4-bx^)P''^dx9 np + fn + 1 np + fn + IJ Each application of formula [B] diminishes j? by unity. Formula [B] fails for the same case as [A], III. To derive formula [C]. Solving formula [A] for CoT'* {a + bsfydx, and substituting m + n for m, we get [C] ffic*** (a + bx^)Pdx a(m + l) a(m + l) J \ t / Therefore each time we apply [C], m is replaced by m -f n. When m-f 1 = 0, formula [C] fails, but then the differential expression can be integrated by the method of § 201, p. 885, and the formula is not needed. IV. To derive formula [!>]. Solving formula [B] for and substituting p -\-l for j?, we get [I>] Cx*^ {a + bx^) P dx ^_xn^^Ha + bxn)P^^ np + n + m + 1 r^n.^a+bxn)P^^dx. Each application of [I>] increases p by unity. Evidently [I>] fails when |? + 1 = 0, but then j? = — 1 and the expression is rational. 348 INTEGRAL CALCULUS 1. r_^^ = _l(je24.2){l-x«)* + C. J Vi — X* ° SohAifm, Here m = 3, n = 2, p = — J, o = 1, 6 = — 1. We apply reduction formula [^i] in this case because the integration of the differential would then depend on the integration of \v,{y — x>)~^(2x, which comes under [4], p. 292. Hence, substituting in [^i], we obtain f-^/t •v-*^ aj»-2+i(l -aJ^)"*"*"^ 1(3-2 + 1) f^ ,,, ^v-4^ = _ 1x2(1 - x»)* 4- f /«{! - xT**B = - }x»(l - x«)* - i(l - x«)* + c = -J(«* + 2)(l-x2)* + C. 2. f-^!^— =-flx8 + ?a»x)V^"^^ + |a*arc8in? + C. •^(a2-x«)* V4 8 / 8 a Exat, Apply [il] twice. 3. r(a« + x«)*dx = ? Va» + x* + ~ log(x + Va* +x2) + C. •/ A 2 ^n<. Here m= 0, n= 2,ps }, a= a*, &« 1. Apply \B\ onoe. . /• dx (X2-1)* 1 ^^ 4. I . = ^ -r- + - arc sec X + C ^ x^yf^^^ 2x3 2 Hint, Apply [C] once. 5. I , = V a^ — X* H — arc sm - + C7. •^ Va« - x» 2 2a. 6. f ^!^ = l(x«-2a2)Vag-f x« + C. •^ voa + xa 3 •^ Vr^ \ 6 16 15/ 8. fx^ Va»-x2dx = ^ (2x2 - a«) Va^ - x» + -arcsm ? + C. •/ 8 8 a mnr. Apply [i<] and then [^. 9. I = h — arc tan - + C. J (a* + xa)a 2a2(oa + x^) 2a» o JERfi^ Apply [/)] onoe. /■ dx _ y/ofi^^^ 1 X ' Jx'VS^rr^a" 2a'^2 ■^2a» ""^a + V^T^a"^^* REDUCTION FORMULAS 849 12. r ^ =(-^^'-^^')%c. 13. r(x« + a2)»da; = 3(2x» + 6a«) Vx^ + a« + ^log(x + Vx^ + a«) + C. 14. rxa(x« + a^^dx = ?(2x» + a«) Vx^ + a^ - ^ log(x + Vx^ + oS) + C. •/ 8 8 16. I = -^- — {2ax-aJ«)* + — arc vers- + C. JWnl. f— ^^=:- fxl(2a-«)-4AF. Apply [^ twice. /dz Va* — X* x»(o«-x«)* o** •^ V2fy-y« ® 2 r 18. f ^ ^ = - (2 at - «*)* + oarc vers - + C. 19. I = 1 arc tan - 4- C J (a2 + a2)8 4o»(aa + ««)» 8a*(as + a'^) 8a» a 20. f ^**^^ = - A (3 ,« + 4r8 + 8) Vl - r« + C. 206. Reduction formulas for trigonometric differentials. The method of the last section, which makes the given integral depend on another integral of the same form, is called 9ucce89ive redv/ition. We shall now apply the same method to trigonometric differentials by deriving and illustrating the use of the following trigonometric reduction formulas. 8in*~x cos** X dx 8in*~ + *a5C08»*-^a5 . n— 1 + 5: r 8in*»*a5C08»*-*a5<to. m + n tn + nj + ^^^^ — I 8in*»-* 0? QOS~ X (fa?, fn + n m + nJ sin*** X cos* X dx 8in'"-^ficco8**+^a5 . f»— 1 860 INTEGRAL CALCULUS [ G] j 8in** X C508* X dx = + — -J 7 — I 8m**a5C508*+*x«a5« n + 1 n + 1 J [ jET] j 8ln'* X C08" X dx ^ 8lii"*^ixco»-^xa, w + n + a r8to«.+.a,co»«a5*D. W»+ 1 t» + 1 J Here the student should note that Formula [E] diminiihes n hy 2. Formula [-F] diminishes m by 2. Formula [O] increases n by 2. Formula [H] increases m by 2. To derive these we apply as before the formula for integration by parts, namely, (A) j udv =zuv— i vdu. (^), p. 841 Let ii = cos"~*a:, and <iv = sin*" 2; cos a;dar; then du = — (n — l)cos''"*a;8ina^a::, and v = sm"''^*2; m + 1 Substituting in {A\ we get (B) I sin"* a: C08"a^ = -f 8in""*"*a;co8"~'a; H :r 1 sin'"'*''a;cos""*a:diB. m4- 1*^ 7n-f 1 n-^1 m-{- In the same way, if we let u = sin'*~^2^ and £2t; = co8"2:sinx(2z^ we obtain /•vx r • « -J sin'""^iccos''"*"*a; ( C) I sm"'ic cos* a:aic = z ^ •/ n 4- 1 m — 1 n + -f 3- I 8in"'"'a:cos"'*"*a;diB. n4- !•/ But J 8in'""*'*a; C08""*a;da; = I sin"*a:(l — cos^x) C08*"'xia? = j sin* a: cos"~* xdx — I sin"'^: cos*xdx. 8in*^a; cos*^a? duo 8in*"+^a5co8*-*a5 . n— 1 REDUCTION FORMULAS 361 Substituting this in (5), combining like terms, and solving for J sin* a: cos'a^rfa:::, we get , H ^T — I 8ln*»a5C08*-*a5cla5. tn + n tn-^nj Making a similar substitution in ((7), we get [F] I sin** as cos* a? ciai; 8in*»*-*a5co8*+^a5 ,«» — !/•*«»• « ^ tn + n tn -^nj Solving formula [E] for the integral on the right-hand side, and increasing n by 2, we get [O] I sin** X cos** X dx »in**+*a5COS**+^x , m + n 4-2 r * ^ «...• ^ = rri . 7^ — I sin*** 35 cos*»+* a? <la?« n+1 n+1 J In the same way we get from formula [F]^ [H] I 8in*'*a5 cos** a? dx 8in***+*a?cos**+^i» «» + n + 2 /• . ^^, ^ , = r-z . J — I sin***+«a5co8**a5<fa?. «» + 1 «» + 1 J Formulas [-B] and [J'J fail when m 4- n = 0, formula [O] when n 4- 1 = 0, and formula [H] when m 4- 1 = 0. But in such cases we may integrate by methods which have been previously explained. It is clear that when m and n are integers the integral J sin"'a:cos''a:di may be made to depend, by using one of the above reduction formulas, upon one of the following integrals: dx r dx /dXf iQinxdxj icoQxdx^ I sin a; cos zdo:, I -: » 1- J J J J sin X J ( sin X •/ cos X dx cos X sin X all of which we have learned how to integrate. : — » I tan xdx^ \ cot xdx^ cos X sm X J */ 852 INTEGRAL CALCULUS /• , • _- J Bin « COB* « . Bin X cob's . 1 . , . v . #• 1. I sin^ X COB* xcix = h + — (Bin x cob x -f «) + C7. J 6 24 10 ■ SolMiion, FirBt applying formula [F], we get (A) JBin^x cos^xdx = h - Jcoa* xdx. [Herem=2, na4.] Applying formula [^] to the integral in the second member of (J.), we get yin /• . , sin X cos* X . 3 f , , (B) I coB^xdx = H - I coB^xdx. CHere m = 0, n B 4.] Applying formula \K\ to the second member of (B) gives sin X COB X X ,r^ /• „ , smxcoBX . X (C) Jco8»xdx = + - Now BubBtitute the result (<7) in (^, and then thiB result In (^i). This gives the answer as above. - C . A o J C08X/8in*x sin'x sinxx . x . ^ 2. I sm*x cos^ xdx = ( ) H h C. J 2\3 12 8/16 3. I = tan X — 2 cot X — cot» x + C. J sin^xcos^x 3 /•cos* xdx cotx.« « 3x ^ 4. I = - -11^(3 - coB»x) - — + C. J BinSx 2 ^ '2 . /•C08*oda COB a 3, ^ o - 6. I — -— — = -r-^ COB a log tan - + C J Bin'a 28inao 1 ^ 2 /. - , COB o /sin* a ,5., .5, \.6a.^ sin* oda = ( sin* a + - sin o I H yC, 2\3 12 8 /16 6 ^ r ^ cobB/ 1 . 3 \ 3, , tf ^ 9. I C0b8 tdt = I COB^ « + - COB* < + — cos* < + — COB O + h C^« J 8\ 6 24 16 /128 10. f ^?? = L_(^_L. + _J 5siny^ •/ sin* ^ cos* y co8*y\3Bin*y 3sin^ 2 / 5 + - log(Bec y + tan y) + C REDUCTION FORMULAS 858 cosnx n 207. To find C^^ sin nxdx and C^^ cob nxda>. Integrating e"* sin nxdx by parts, letting w = e"', and dv = sin nxdx; then du = ae*^dx^ and t;= — « Substituting in formula (-4), p. 841, namely, we get / >i\ r ax ' J ^"^ cos wx . a /* „ , (-4) I e" sm wzaa: = h - I « cos nxos; •/ n nJ Integrating «** sin n2:da; again by parts, letting u = sin nxy and dv = e'^dx; then c7if = n cos nxdx^ and v = — • a Substituting in (A)^ p. 341, we get {B) J «"* sin warda: = ^ I e"* cos nxdx. Eliminating I e"* cos nxdx between {A) and (5), we have (a^ 4-71*) I g" sin nxdx =. e'^{a sin nx — n cos na;), /' . , e'^ia sin na: — w cos na;) ^ e*" sm narda; = — ^^ '- -f C. or -f n* Similarly we may obtain /' , 6"'(n sin na; 4- a cos na;) . ^ «"* cos wa;ia; = — ^^ , ^ ^ '- 4- C. In working out the examples which follow, the student is advised not to use the above results as formulas, but to follow the method by which they were obtained. 854 INTEGRAL CALCULUS e* COB xdx = — (sin x + cos x) + C7. 2 3. fe^'cosSxdx = — (Sein 3x + 2 cob3x) + O. . /"sin xdx sin X + cos x . ^ 4. I = ^. c. 6. I = — — -(28in2x-3coB2x) + C. J c»' 13c»* A r • • « ^ ^/t 2 8in2x + co8 2x\ . -, 6. I c*8in* xdx = — ( 1 ^ ) + C. n r « J e«/- . 2sm2o + cos2o\ . -- 7. J c*cos9ada = — f 1 + -^^ j + 0. a rc^co8-dx = e«^8in- + coe5^ + C. J 2 • \ 2 2/ c^^sinott 9. I eo^(8Uiaa + cosaa)(ia = h C 10. reS''(Bin2x-C08 2x)(ix= -^(8in2x ~6cob2s) + C. •/ 13 CHAPTER XXIX --a-4 M N THE DBFINITB DTTBGRAL 208. Differential of an area. Consider the continuous functioA ^(a?), and let be the equation of the curve AB, Let CD be a fixed and MP a variable ordi- nate, and let u be the measure of the area CMP J).* When x takes on a suf- ficiently small increment Ax, u takes on an increment At^ (=:area MNQP). Completing the rectangles MNRP and MNQS^ we see that area MNRP < area MNQP < area MNQS, or, -MP.Ax< Aw<i\rQ.Ax; and, dividing by Ao;, Al£ Now let Ax approach zero as a limit; then since MP remains fixed and NQ approaches MP as a limit (since y is a continuous function of x), we get du = V(=MP), or, using diflferentials, dx du = ffdx. Theorem. ITie differential of the area bounded by any curve^ the axis of X^ and two ordinates is equal to the product of the ordi- nate terminating the area and the differential of the corresponding abscissa. • We may suppose this area to be generated by a yariable ordinate starting out from CD and moTing to the right ; hence « will be a function of x which yanishes when x^a. t In this figure MP is less than NQ; if MP happens to be greater than NQ^ simply reverse the inequality signs. 865 866 INTEGRAL CALCULUS 309. The definite integral. It follows iiom the tlieorem in the last section that if AB is the locus of then du = ydr, or, (A) du=^{x)dx, where du is the differential of the area between the curve, the axis of x, and any two ordinates. Integrating {A)^ we get w = \^{x)dx. Since \^{x)dx exists (it ia here represented geometrically as an area), denote it hyf(x)+ C. (B) .■.u=f(T) + C. We may determine C, as in Chapter XXV, if we know the value of u for some value of j^ If we agree to reckon the area from the axis of y, i.e. when (C) x = a, u = eLre&OCDG, and when x=b, u = area OEFG, etc., it follows that if (D) x=% then m = 0. Substituting {D) in {B), we get w=/(0)+C, or C = -/(0). Hence from (B) we obtain {E) u=f(z)-f{0), giving the area from the axis of y to any ordinate (as MP). To find the area between the ordinates CD and EF, substitute the values (C) in {E), giving {F) area OCDG =f{a) -/(O), (Q) area OEFG =/(6) -/(O). Subtracting (F) from (G), (H) area CEFD = f{b) - /(«).■ > The itHdent fnticllOD ifUcIi d: THE DEFINITE INTEGRAL 357 Theorem. The difference of the value% of j ydx for a; = a and x=b gives the area bounded hy the curve whose ordinate is tfy the axis of Xy and the ordinates corresponding to x=a and x = b. This difference is represented by the symbol* (/)" I ydx, or i (f>(x)dx, •/o •/a and is read " the integral from a to ( of ydxJ*^ The operation is called integration between limits, a being tlie lower and b the upper limit.f Since (J) always has a definite value, it is called a definite integral. For, if f<f>{x)dx=f{x)-^C, then £<t>{x)dx = [f(x)-^c][=[f(b) + c]-^[f(a) + c], . or, £<l>{x)dx=f{b)^f{a), the constant of integration having disappeared. We may accordingly define the symbol I <f>(x)dx as the numerical measure of the area bounded by the curve y =^<\> (:r), J the axis of X, and the ordinates of the curve at x=:a, x=b. This definition presupposes that these lines bound an area, i,e. the curve does not rise or fall to infinity, and both a and b are finite. We have shown that the numerical value of the definite integral is always /(6) —f{a), but we shall see in Ex. 2, p. 363, that/(6) —f{a) may be a number when the definite integral has no meaning. 210. Geometrical representation of an integral. In the last section we represented the definite integral as an area. This does not necessarily mean that every integral is an area, for the physical interpretation of the result depends on the nature of the quantities * This notation Is due to Joseph Fourier (1768-1830). t The word limit In this connection means merely the value of the varlahle at one end of its range (end value), and should not he confused with the meaning of the word in the Theory of Limits. $ ^ (a;) is oontinaouB and single-ralued throughout the interral [a, b]. 858 INTEGRAL CALCULUS represented by the abscissa and the ordinate. Thus, if x and y are considered as simply the coordinates of a point and nothing more, then the integral is indeed an area. But suppose the ordi- nate represents the speed of a moving point, and the corresponding abscissa the time at which the point has that speed; then the graph is the speed curve of the motion, and the area under it and between any two ordinates will represent the distance passed through in the corresponding interval of time (see § 187). That is, the number which denotes the area equals the number which denotes the distance (or value of the integral). Similarly a definite integral standing for volume, surface, mass, force, etc., may be represented geometrically by an area. On page 872 the algebraic sign of an area is interpreted. 211. Mean value of f(x). This is defined as follows : r Q^ L n ^^^^^^ ^r } / ^ M A c B ar* a Z' "h sy Mean value of f(x) ^ j^r^^)^ from a5 = atoa5 = 6)"" 6 — a Since from the figure J <l>(x)dxz= area AFQB, this means that if we construct on the base AB(=b^a) a rec- tangle (as ALMB) whose area equals the area of AFQB, then , 2iVQ2i,ALMB AB'CR ,^.^ , ^^ mean value = — = — — = altitude CB. b — a AB 212. Interchange of limits. Since J <\>{x)dx=f{b)—f(a)^ and we have £<t>{x)dx=f(a)-f(b) [/(6)-/(a)], . j ^{x)dx=—i ^{x)dx. Theorem. Interchanging the limits is equivalent to changing the sign of the definite integral. THE DEFINITE INTEGRAL 359 213. Decomposition of the interral of integratloa of the definite int^ral. Since J<(>{x)dx=/(x{)~f{a) and f,i>{x)dx^f{b)-f{x;), we get by addition, fy{x)dx+f4.{x)dx=f(b)-f(a). But j^<f, (x) dx =f{b) -f{a) ; therefore by comparing the last two expressions we obtain Interpreting this theorem geometrically, as in § 209, p. 356, we see that the integral on the left-hand side represents the whole area CEFD, the first integral on the right-hand side the area CMPD, and the second integral on the right-hand side the area MEFP. The truth of the theorem is therefore obvious. Even if x^ does not lie in the interval between a and h, the truth of the theo- rem is apparent when the sign (p. 372) as well as the magnitude of the areas is taken into account. Evidently the definite integral may be decomposed into any number of separate definite integrals in this way. 214. The definite integral a function of its limits. From £4,(x)dx^f(b)-f{a) we see that the definite integral is a function of its limits. Thus I ^ (z) dz has precisely the same value as I <ft(x)dx. Theorem. A definite irUegral is a Junction of its limitt. 860 INTEGRAL CALCULUS 215. Calculation of a definite integral. The process may be sum- marized as follows : First step. , Find the indefinite integral of the given differential expression. Second step. Svbstitute in this indefinite integral first the upper limit and then the lower limit for the variable^ and subtract the last result from the first. It is not necessary to bring in the constant of integration, since it always disappears in subtracting. Ex. 1. Find r z^dx. 1 L3 J 1 3 3 Ex. 2. Find f Bin xdz. Jo Solution. C Bin xdx = f - cob x 1 = I — (- 1) I — — 1 = 2. Aia, ,'* dx Ex. 3. Find ' J'" ox a« + x« J"" dx rl x"i *• 1 1 = - arc tan - = - arc tan 1 — arc tan a*-* + x^ La a J o a a *" ^ IT . = = — jing, 4a 4a 216. Infinite limits. So far the limits of the integral have been assumed as finite. Even in elementary work, however, it is some- times desirable to remove this restriction and to consider integrals with infinite limits. This is possible in certain cases by making use of the following definitions. When the upper limit is infinite, and when the lower limit is infinite, £j> (X) dx = „ ^^X fj (^) <i^ provided the limits exist. THE DEFINITE INTEGRAL 861 Ex. 1. Find f^*^. Solution. r^= li^^J' r^^^ limit r^ll* Ex. 2. Find I . 5oZu<io7i ^"^ i +• Sa*dz +* 8a«dg ^ limit f^ Sa^cte __ Umit X 1* " arc tan — 2aJo = .^^™!^^r4aaarctanA"|=:4aa.!: = 2Ta». ulna. = +QoL 2oJ 2 Let OS interpret this result geometrically. The graph of our function is the witch, the locus of 8a8 y =z . x«-+-4a« Area 0Pe6= f^-^^^^ = 4 a^ are tan A . Jox^ + ^a^ 2a Now as the ordinate Q6 moves indefi- nitely to the right, & 4 a^ arc tan is always finite, and limit r4a2aretanA] = 2Ta2, 6 = +ooL 2aJ 2a 6 >4-ao which is also finite. In such cases we call the resuU the area bounded by the curve^ the ordinate OP, and OX, although strictly speaking this area is not completely hounded. Ex. 3. Find C +* dx X SoltUion. + «dx /•+«ax__ limit r'*ax_ limit ..^„j^. Jl 7""& = + ooJi •^-6 = +oo^^^«^^- The limit of log & as 6 increases without limit does not exist ; hence the integral has in this case no meaning. 217. When ^(o?) is discontinuous. Let us now consider cases when the function to be integrated is discontinuous for isolated values of the variable lying within the limits of integration. Consider first the case where the function to be integrated is continuous for all values of x between the limits a and ( except x = a. 862 INTEGRAL CALCULUS If a < 6 and e is positive, we use the definition (A) f4>ix)dx=^^^lf<l>{x)dx, and when </> (x) is continuous except at a: = 6, we use the definition (B) fj (x) dx = ;'^'^ jr'" V ix) dx, provided the limits are definite quantities. Ex. 1. Find C dz Vaa-x« Solution. Here — — =: becomes infinite for x = a. Therefore, by (B), r «_^_ ^ limit r— _dx_ ^ limit T^^ ^j^ xl — ^0 Va2 - x2 « = 0*^o Va^ - x^ « = 0L aJo = ]^™>^rarc8ln(l-')] = arc8inl = ^. ^n«. J •Idas x2 Solution. Here — becomes Infinite for x = 0. Therefore, by {A), x* Jo x^ e = Ojc x2 e = 0\« / In this case there is no limit and therefore the integral does not exist If c lies between a and b and <f>{x) is continuous except at 2;=:<?, then, € and e' being positive numbers, the integral between a and J is defined by (C) £<l>{x)dx='^^'l£"4>(z)dx + ^''^l£j(x)dx, provided each separate limit is a definite quantity. Ex. 1. Find /•»« 2xdx (x2 - a2)i Solution, Here the function to be integrated becomes infinite for x = a, i.e. for a value of x between the limits of integration and 8 a. Hence the above defini- tion (C) must be employed. Thus, /•««_2xdx_ _ limit r°-* 2xdx limit /•**• 2xdx =«^'^[3(.-a,»];-v»-"o[«(*'-«')»]::.. = J*So[3-^(a-.)=-a« + 8a']+ ^'^'* [S-I^^ - 3 v'(a + 0«-«'] = 3o' + 6a' = 9o*. An$. TUE DEFINITE INTEGRAL To interpret thU geometrically, let us plot the gnph, i.e. tbe locua of = 3V(o-«}*-a* + 3a'. Now u PE movee to the right tonards the asymptote, Le. m « approache* zero, is always finite, and ™i[a^ which U also flaite. As in Ex. 2, [ the asymptote, aod OX. Similarly, S61, 8 a* is called the area bounded by OP, nE'QRO is always finite as QE" moves to the left towards the asymptote, and as ^ approaches zero the result 6 a' is also finite. Hence 6 a' is called the area between QR, the asymptote, the ordinate x = Sa, and OX. Adding these results we get 9a}, which is then called tbe area to the right of OT between the curve, t^e ordinate x s 3a, and OX. Ex. 2. Find /■''' dx Jo (»-a)a' Solution. This fancllon also becomes infinite between the Umita of integraUon. Hence, by (C), ^ ^ ,,_^,^ ^ ^^^^ d, Jo (x-a)» » = 0j) (i-o)' •' = OJ-+.'(J!-a)» ^ _ llmitr 1 T- ■ limit T 1 1*" _liniit/l 1\ , limit/ 1 . 1\ In this case tbe limits do not exist and the integral has no meaning. If we plot the graph ol this function and note the limits, the condition of things appean very much the same as in the last example. It turns out, however, that the shaded portion cannot be properly spoken of as an are&, and the integral sign has no meaning in this case. That It Is important to note whether or not the given function becomes Infinite within the limits of Integration will appear at once If we apply our Inlegtation formula without any investigation. Thus, Jt ix-a)* L !r-aj« a result which ia absurd In view of the above discnssious. 864 INTEGRAL CALCULUS 1. r*6x%ix = 38. 2. r (a«x-x«)dx = - 3. r^=i. '(2x J "ax — = L 1 X 5. C^ (xa - 2x + 2)(x - l)(ix = •^° V3-2X -f .2x«dx 8 7. j - = --log3. Jo X + 1 3 dz V2 - 3x2 4V3 9. f' f^ =-^126. 2-v^?^ Jo yS - y + 1 3 V3 r^ tdt _ log 2 ' J2 1 + ia - 2 Jl X2(l-f + * dx + = l-log2. Jo a^ + x^ 2a' ji «(i+«)^ '^ 2 5. r* sin 0d^ = 1. J*2ir er8aftW = 3x. 7. p8ec*^(W = f. J '•I T arc sin xdx = 1. 2 L Jj'xlog xdx = e^ + l 20. r e-«dx = l. 21. f arc tan xdx = - — log Vi. Jo 4 /•s*" V2r 22. J -4r-dx = 4r. vs 23 • j^*(fVt-A«»)d« = 2V6-6. 24. r /^ =^. 26. r^;^=8r. •^0 V2r-y 256 IT 6* 27. 2a r (2 + 2co8^)*d5 = 8a, 28 . r* sin* a cos* ada = ^^j. 29. r*^tanoda = 0. 30. r log ydy = - 1. 3L J xlogxdx = — J. 82. j;'xnog«d« = -t. 33 •X a^ sin ada = ir — 2. 34. r^8ec^ = log(?-tp:-?) 6 ^0 2 + cos a 3 V3 36 J '2 cos Q^ __ X l + 8in2tf ~4 THE DEFINITE INTEGRAL 865 218. Change in limits corresponding to change in the variable. When integrating in Chapter XXVII by the substitution of a new variable it was often found quite troublesome to translate our result back into the original variable. When integrating between limits, however, we may avoid the restoration of the original variable by changing the limits to correspond with the new variable.* Ttiis process will now be illustrated by an example. Ex. 1. Find C Va* — x' (2x, assuming x = asin $. Solution. Va^-x^dx = a* Vl - Bin^ e . cos Bde = o«co8« 6d$. When x = a, a = a^in9f i.e. 9 = -; and when x = 0, = a sin 9, i.e. $ = 0. .•.j(;'VS?T^<te=j;i«*co8«ftl»=[|(» + !!!^)]» = ^. An,. By this change in limi^ we mean that as $ increases from to - « x will increase from to a. EXAMPLES 1. r - = 4-21og3. Assume Vx = «. •^M + Vx a dx Assume x^az. 2. f ^ „ 8. r'<^-)^ = 6. . A««me* = l. Ji X* z 4. r*__^_ = 1. Assume VT^ = 2. •^0 Vl -x« 5. r*sinaco8«o(ia = i. Assume sin a = «. g r?(Bin ^H- cos g)(i^^ logs Assumesin^-cos^c=z. Jo 3 + sin 2^ 4 7. I = arc tan c — - • Assume c* = «. Joe*-\-e-' 4 8. r°_ ^g — _ ^ Assume x = o sln*«. •'o Vox-xa • The relation between the old and the new variable should be euch that to each ralue of one within the limits of integration there is always one, and only one, finite ralue of the other. When one is given as a many-vained function of the other, care must be taken to choose the right values. 366 INTEGRAL CALCULUS 3V3 9. r<^-"i'^=8+ J^ (ai-2)J + 3 Aasame x — 2 = z*. — dx = 4 — r. Assume e* - 1 = z*. c* + 3 1. r -J^_ = i^. Assume 2 + 4y = z«. ^i V2 + 4y 2 Assume 2 = tan - • 2 cos t V5 2 Jo 3 + L r — dx = 4 — T. Assume c* — 1 = «*. Jo c* + 3 , /•« x^dx 8V2-4 . L I 5 = a. Assume x = a tan 2. Jo (a2 + x*)* 2 J* * y irci* y2 Va* — y^dy = — . Assume y = a sin z. *" '^ *" 16 6. r V2« + t2(tt = V3- Jlog(2+V3). Assumet + 1 = «. J«»l0J« 4 — IT V c* — 1 dx = Assume e* + 1 = «. 2 8. I — do = - log 3. Assume sm — cos = z, Jo 3 + sin2^ 4 /.i+v^ (x» + l)dx , . .1 9. I — = log 3. Assume x = «. •^^ xVx*4-7xa + l * CHAPTER XXX INTEGRATION A PROCESS OF SUMMATION 219. Introduction. Thus far we have defined integration as the inverse of differentiation. In a great many of the applications of the Integral Calculus, however, it is preferable to define Integra^ tion as a process of summation. In fact the Integral Calculus was invented in the attempt to calculate the area bounded by curves by supposing the given area to be divided up into an "infinite number of infinitesimal parts called elementSy the sum of all these elements being the area required." Historically the integral sign is merely the long /S, used by early writers to indicate " sum," This new definition, as amplified in the next section, is of fun- damental importance, and it is essential that the student should thoroughly understand what is meant in order to be able to apply the Integral Calculus to practical problems. 220. The definite integral defined as the limit of a sum of differ- ential expressions. Assume (f>{x)dx as the differential of f{x). Then by § 209, p. 366, (^) f<l>{x)dx=f{b)^f{a) gives the area bounded by the curve y = <f>{x) (AB in figure), the axis of X, and the ordinates 2; = a and x = b.* Now suppose the segment ai to be divided into a number of equal parts, say 6, each equal to Ax^ at points whose abscissas are b^^ b^ 6„ b^^ b^. Erect the ordinates at these points and apply the Theorem of Mean Value (44), p. 168, to each interval, noting that here <\>{x) takes the place of 4>\o^, r • In the figure a, 6^ &,, x,, Xfi, etc., denote the abfclMM of the points nnder whidh thej Are written. 867 368 INTEGRAL CALCULUS Applying (44) to the first interval (a = a, 6 = 6i, and x^ lies between a and b^ as shown in figure), we have 0^ — a or, since 6^ — a = Aa:, f{b;)-f{a)=<f>{x,)Ax. Applying (44) in the same way to each one of the remaining five segments, we get /(!>*) -f{f>^=4>{^.)^ f{K)-nh)=H^.)i^ m)-m)=<^(^.)^ . , f{b)-f{b,)=4>(x,)Ax, respectively. Adding these six equations, (B) f{b) -/(a) = </> (X,) Aa: + </> (x,) Ax + </> (x,) Ax + ^ (ar J Aa: 4- </) (Xg) A2: + 4>{x^)Ax. But <l>{x^)Ax = area of first rectangle, ^(z2)Ax= area of second rectangle, etc. Hence the sum on the right-hand side of (B) equals the area of the figure bounded by the zigzag line BF^p^P^p^^p^P^p^F^p^Q^ the axis of Xj and the ordinates x=:a and 2; = (. It is also evident that the area of this figure equals f(f>) -/(«), no matter into how many equal parts the interval [a, 6] may be divided. Hence for any number n of equal parts (0) f{b)^f{a) = <l>{x,)Ax-^<f>{x,)Ax-^...+<f>{x,)Ax, IB) where Ax=z "" . n When the number of segments (=n) into which the interval [a, 6] is divided increases without limit, equations {(T) and (2>) still hold true, but Ax becomes dx^ i.e. a variable whose limit is zero (§30, p. 21). INTEGRATION A PROCESS OF SUMMATION 869 .-. /(J) ~f{a) = ^ta'^ [.^ (^,) i:r + ^ (^ ir + . - . + ^ (^.) dx}, or, by {A), (E) £4>{x)dx^^^fl{.^{x,)dT+4,{x,)dx^.-. + <^{x,)dx\ This exhibits our definite integral as the limit of a sum of differ- ential expresaiona. Each one of the differential expressionB ^(a',)(2x, ■ ■■t 4* (^i) ^^ '3 called an element of the integral. The above result also illustrates very clearly the definition of the definite integral as the area under the curve. For as n increases without limit the sum 4,(x^)dx + <t){x;)dx + ---+<f>(xjdx always represents the area under the zigzag line, and evidently the figure bounded by the zigzag line, the end ordinates, and OX approaches the area under the curve as a limit. We may, however, attack the problem in a much more general way, for it is geometrically evident that a Bubdivieion of the given area may be made in an infinite variety of ways such that the con- tinuation of the process will lead in the limit to the area desired. For example, let us choose within the interval [a, 6], n — 1 abscissas, Zi, x^ '■-, x^_^, in any manner whatever, and erect ordinates at the corresponding points to the curve. Then the area is divided into n portions such as x^P, x^Q, ete. Denote the lengths of the sub- divisions on OX as follows : x,-a = Ax^, Xt-Xi = Aa^ a:,_, — a:,_, = Ax,_„ h — a:,_, = Ai.. Then evidently the area of the segment x^Q, for example, equals approximately that of the rectangle whose base is Az^ and altitude Qx„ or (x,). Carrying out this idea for each portion, we have as a result that *(ar,)Ar, + 0(s,)Az, + ..-4.0{i._,)Ax„, + 0(OA3:J gives an area approximately equal to the area required. And now it is geometrically obvious that the limit of the above sum is the 370 INTEGRAL CALCULUS area under the curve if the process of subdivision of [a, b] be continued according to some law by which each division on x approaches the limit zero. That is, ^=lX<t>i^d^,*=£4>(=ddx. This general discussion shows that inteffration between limits is a process of summation^ in which, however, the definite integral appears not as a simple sum but as the limit of a sum^ the number of terms increasing without limits and each term separately approach- ing the limit zero. In order to replace the intuitional point of view that we have so fax adopted in the text by a rigorous and general analytical proof, proceed as follows. Divide the interval [a, 6] into any number of parts Axi, Axs, • -*, Az., and let vfi be any abscissa in the segment Axi, the extremities of this segment being denoted by X{ and Xi^ i. Also suppose Xj to be a value in the interval [x„ Xi^i\ determined by the Theorem of Mean Value (as in the first part of this section), so that /(X.+ 1) - /(Xj) = (x;) Ax,-. Then, exactly as hrfoTBy the sum equals the required area. And while the corresponding sum n (O) ^^{x'i)Axi 1 does not also give the area, nevertheless we may show that the two sums (F) and (G) approach equality when n increases without limit. For the difference 4>{xi) — (z'l) does not exceed in numerical value the difference of the greatest and smallest ordinates in Ax^. And furthermore, it is always possible t to make all these differ- ences less in numerical value than any assignable positive number e, however small, by continuing the process of subdivision far enough, Le. by choosing n sufficiently large. Hence for such a choice of n the difference of the sums (F) and {G) is less in numerical value than e (& — a), i.e. less than any assignable positive quantity, however small. Accordingly as n increases without limit, the sums (F) and (O) approach equality, and since (F) is always equal to the area, the fundamental result follows that ^ n in which the interval [a, b] is subdivided in any manner whatever, and x'j is any abscissa in the corresponding subdivision. * Any term of this sum is Bometlmee called an element of the area, for each one represents the area of one of the rectangles forming the whole figure. t That such is the case is shown In advanced works on the Calculus. INTEGRATION A PROCESS OF SUMMATION 871 The process of evaluation of the limit of this sum is accordingly often spoken of as summing up an infinite number of infinitely sm^ill quantities^ but the phrase has no meaning except in the above sense. We may apply to great advantage this fruitful idea of summing up an infinite number of infinitely small quantities to a large number of the problems of the Integral Calculus. In order to obtain solu- tions by this method the following steps are in general to be taken. First step. Find a differential expression for any one of the infin- itesimal quantities (elements) composing the quantity to be calcu- lated,* and reduce it to a form involving only a single variable. Second step. Integrate this differential expression (i.e. sum up all the elements) between the limits given by the conditions in the problem. 221. Areas of plane curves. Rectangular coordinates. It was shown in § 209, p. 856, that the area between a curve, the axis of X and the ordinates x = a and x=b is given by the formula U) area = J ydXi the value of y in terms of x being sub- stituted from the equation of the curve. Considered as a process of summation, it is customary to look upon this operation as follows (see p. 867). Consider any strip (as CE) as an element of the area. Regard- ing it as a rectangle of altitude y and infinitesimal base dx, its area is equal to ycZx, and summing up all such strips between the ordi- nates AF and BQ gives the area ABQP. Similarly it may be shown that the area between a curve, the axis of F, and the lines y = c and y = rf is given by the formula (B) area =X' xdy^ the value of x in terms of y being substi- tuted from the equation of the curve. * If an area la wanted, find /i differential exprecMion for an element of the area ; if a length, fled it for an element of the length ; if a volume, find it for an element of the Tolume, etc. 372 INTEGRAL CALCULUS Ex. 1. Find the area included between the semicubical parabola y< = x* and the line x = 4. SoliUioTL Let us first find the area OMP^ half of the required area OPP'. For the upper branch of the curve y = Vx", and summing up all the strips between the limits x = and x = 4, we get, by substituting in (A), / area OMP =Cydx -Cz^dx = V = ^^' Hence area OPP" = 2 . V = 26§. If the unit of length is one inch, the area of OPP is ^^X 26 J square inches. Note. For the lower branch y = — x* ; hence area OMP" = r(- x^) dx = - 12f . This area lies below the axis of z and has a negative sign because the ordinates are negative. In finding the area OMP above, the result tooa positive because the ordinates were positive, the area lying above the axis of x. The above result, 25 j^, was the total area regardless of sign. As we shall illus- trate in the next example, it is important to note the sign of the area when the curve crosses the axis of X within the limits of integration. Ex. 2. Find the area of one arch of the sine Y curve y = sin x. Solution. Placing ^ = and solving for x, X = 0, IT, 2 IT, etc. Substituting in (A), p. 371, we find Also, and area OAB = C ydz = j sin xdx = 2. area BCD = C ydz = ( 'sin xdx = — 2, area OABCD -C ydz = f 'shi xdx = 0. This last result takes into account the signs of the two separate areas composing the whole. The total area regardless of these signs equals 4. Ex. 3. Find the area included between the parabola x^ = 4 ay and the witch 8o» y = x2 + 4 a2 Solution. To determine the limits of integration, we solve the equations simul- taneously to find where the curves intersect. The coordinates of A are found to be (-2 a, a), and of C (2 a, a). INTEGRATION A PROCESS OF SUMMATION 373 It is Been from the figure that area ^OCB Bat BxeADECBA and KteskDECOA Hence «re&AOCB ATeskDECBA - areaDJSCO^. •»« SaMx = 2Ta' xa + 4a« 2«x2 , 4a2 = 2 X area 0-EC = 2 f — dx = Jo 4a = 2ira« - — = 2a2(T - f). Ans, Another method is to consider the strip PS as an element of the area. If y' is the ordinate corresponding to the witch, and y^' to the parabola, the differential expression for the area of the strip PS equals (y^ — y'^dz. Substituting the values of 2^ and y'' in terms of z from the given equations, we get Y axeskAOCB = 2 x area OCB = 2rV-y")<^ Jo \x2 + 4a« 4a/ = 2a»(T~i). Ex. 4. Find the area of the ellipse — + ^ = 1. Solution. To find the area of the quad- rant OAB, the limits are x = 0, z = a ; and y = - Va2 - x2. a Hence, substituting in (^), p. 371, area 0-4B = - Cia^ - x2)*dx aJo = — (o^ — a;2)> + arc sin L2a^ '2 oJo ra6 Therefore the entire area of the ellipse equals wab. 222. Area when curve is given in parametric form, equation of the curve be given in the parametric form z=/(0, y = <f>{t). [B], p. 346. Let the 874 INTEGRAL CALCULUS We then have i/=:<f>(t) and dx =f'{t) dt, which substituted* in (-4), p. 371, gives (A) area=: C*^(t)f^{t)dt^ where t = t^ when a? = a, and t = t^ when a; = J. We may employ this formula {A) when finding the area under a curve given in parametric form. Or we may find y and dx from the parametric equations of the curve in terms of t and dt and then substitute the results directly in (A)^ p. 371. Thus, in finding the area of the ellipse in Ex. 4, p. 373, it would have been simpler to use the parametric equations of the ellipse X = a cos 0, ^ = 6 sin 0, where the eccentric angle ^ is the parameter ($ 79, p. 94). Here y = 6 sin and dx = — a sin 0(2^. When X = 0, = - ; and when x = a, = 0. Substituting these in (A)^ p. 371, we get areaOJ.B=l ydx = — J a6sin*0<i0 = -— . 1 Hence the entire area equals rob, Ans, EXAMPLES 1. Find the area bounded by the line y = 6x, the axis of X, and the ordinate x = 2. Am. 10. 2. Find the area bounded by the parabola |/< = 4 x, the axis of X, and the lines X = 4 and x = 9. Ana, 2b\* 3. Find the area bounded by the parabola y' = 4x, the axis of F, and the lines y = 4t and y = 6. Ana. 12}. 4. Find the area of the circle x' + y* = r^. Ans. rr*. 5. Find the area between the equilateral hyperbola xy = a>, the axis of X, and the ordinates x= a and x = 2 a. Ana. a31og2. 6. Find the area between the curve y = 4 — x^ and the axis of X. Ana. 10]. 7. Find the area intercepted between the coordinate axes and the parabola X* + y* = a*. s Ana. _. 6 • For a rigorouB proof of this satMtltutlon the student is referred to more advanced treatises on the Calculus. INTEGRATION A PROCESS OF SUMAIATION 375 8. Find the area bounded by the semicubical parabola y' = z\ the axis of F, and the line y = 4. ^„^, J 4^1024. 9. Find the area between the catenary y=- ^ + e ^ L the axis of F, and the line x = a. *^ "- - a*ro ^-. 10. Find the area between the carve y = logx, the axis of JT, and the ordinates X = 1 and X = a. Ana, a (log a — 1) + 1. 11. Find the entire area of the curve (ly- (D - '■ 12. Find the entire area of the curve a*y* = x» (2 a — x). Ans. ira*. 13. Find the area bounded by the curves X (y — e*) = sin X and 2 xy = 2 sin x + x*, the axis of F, and the ordinate x = 1. Ana. (^(e* — }x^dx = e— { = 1.56 + • • •. 14. Find the area between the witch y = —„ — j— « smd the axis of X, its X* -f- 4 cfl asymptote. Ana. ^ra^, x* 15. Find the area between the cissoid y* = ^ and its asymptote, the line X = 2 a. Ana. 3 ra^. 16. Prove that the area bounded by a parabola and one of its double ordinates equals two thirds of the circumscribing rectangle having the double ordinate as one side. 17. Find the area included between the two parabolas v^ = 2px and x> = 2py. Ana. — ^' 8 18. Find the area included between the parabola y^ = 2x and the circle y« = 4x-x2. Ana. 0.476. 19. Find the total area included between the curve y = z^ and the line y = 2x. Ana. 2. 20. Find an expression for the area bounded by the equilateral hyperbola z^ — y^ = a^, the axis of X, and the diameter through any point (x, y). An;a. — log • 2 * a 21. Find by integration the area of the triangle bounded by the axis of Y and the lines 2x + y + 8 = and y = — 4. Ana. 4. 22. Find the area of the circle IX = r cos By y = rsintf; $ being the parameter. Ana. in4. 876 INTEGRAL CALCULUS 23. Find the area of one arch of the cycloid (x = a(tf-8intf), \y = a(l — CO80); $ being the parameter. Hint. Since x vorieB from to 2 wa, varies from to 2 a-. Aris. 3 ra^ ; that is, three times the area of the generating circle. 24. Find the area of the hypocycloid IX = o cos* Of y =:a sin* $ ; $ being the parameter. T Ara, 8 area of the circumscribing circle. ; that is, three eighths of the 25. Find the area of the loop of the folium of Descartes sc' + y» = 3 axy. 3ai Bint. Let y^.tx; then X' l + /» 1-f ^ The limits for t are and oo. y=l<!^,anddir-l:i2^3c««. {1 + !»)« 223. Areas of plane curves. Polar coordinates. Let J^Cbeacurye whose equation is given in polar coordinates. Let (/>, 0) be the coordinates of P, and assume u as the measure of the area OEPJ* When takes on a small increment A5, u takes on the increment Am(= area OPQ). Completing the circular sectors OPR and OSQ^ it is seen that C F or area OPR < area OPQ < area OSQ, Dividing through by A0, Au Now let A0 approach zero as a limit ; then OQ will approach OP as a limit, and we get dd 2^\ 2 J * Since we may enppofle this area to be generated by a variable radliu reotor starting out from OE and moving up to the position OP, u will be a function of 9 which vanishes when 9s a. t The area of a circuliu' sector = i radius x arc= ^ OP x 0P^6. t In this figure OP is less than OQ ; if OP is greater than OQ, simply reverse the Inequality signs. INTEGRATION A PROCESS OF SUMMATION 377 Or, using differentials, du= ^ p'dO, this being the differential of the area in polar codrdinatet. Integrating, we have u = ^ fp'dO. If we now apply the same line of reasoning as that followed in § 209, p. 856, the area of the sector OEF may be calculated by s of the formula .=!jVd», the value of ^ in terms of being substi- tuted from the equation of the curve. To look at this process as a summation, consider any sector OBC as the element of area. Regarding it as a circular sector of radius p and infinitesimal arc pd0, its area equals ^ p*d0. Summing up all such sectors between OE and OF gives the area OBF. Formtda (A) may then he uted for finding the area hounded hy a polar curve and ike radii vectors correaponding to 6^a and = p. Ex. 1. Find the entire area of the lemniscate p' = <ii cos 2 8. Solution. Since the figure la symmetrical with respect to both OX and OF, the whole PiP,ir) area = 4 times the area of OAB. Since p = when ^ = j> ve see that it 9 varies from to - • the radius vector OPsweeps OAB. Hence,snlMtitutingiD(^), 'tkOAB = 4.iJ pW = 2a'J'*cqB2AM = o»; a of a square constructed on OA as 1. Find the area swept over in one revolution by the radius vector of the spiral of Arcbiinedee, p = a$, starting with fl = 0. . 4 ir*a' 3 2. Find the area of one loop of the curve /> = ocoa 2 A ^^ ^. 878 INTEGRAL CALCULUS 3. Show that the entire area of the curve p = a sin 2 tf equals one half the area of the circumscribed circle. 4. Find the entire area of the cardioid p = a (1 ~ cos ^). Ana. ; that is, six times the area of the generating circle. 5. Find the area of the circle p = a cos tf. . xa* jn.n8, • 4 6. Prove that the area of the three loops of p = a sin 3 equals one fourth of the area of the eircumscribed circle. 7. Prove that the area generated by the radius vector of the spiral p = e* equals one fourth of the area of the square described on the radius vector. 8. Find the area of that part of the parabola p = a sec^ - which is intercepted between the curve and the latos rectum. . 8 a^ Ana, — . 3 9. Show that the area bounded by any two radii vectors of the hyperbolic spiral p0 = a ia proportional to the difference between the lengths of these radii. 10. Find the area of the ellipse p8 = — — — — Ana, wab. '^ a2 sin* e-\-l^ cos2 11. Find the entire area of the curve p = a (sin 2 + cos 2 0), Ana. ira\ 12. Find the area of one loop of the curve p> cos 9 = a> sin 8 ^. «, « « - 3 a* o*. _ Ana, -— -T-log2. 13. Find the area below OX within the curve p = a sin> - . Ana. (10ir + 27V3)^- 224. Length of a curve. By the length of a straight line we commonly mean the number of times we can superpose upon « it another straight line employed as a unit of length, as when the carpenter measures the length of a board by making end-to- end applications of his foot rule. ^^^^^^»w^^ Since it is impossible to make a j^^^ ^^\. straight line coincide with an arc of a Y ^1^ curve, we cannot measure curves in the I same manner as we measure straight lines. We proceed then as follows. Divide the curve (as AB) into any number of parts in any man- ner whatever (as at C, i>, E) and connect the adjacent points of division, forming chords (as AC, CD^ DE^ EB). INTEGRATION A PROCESS OF SUMMATION 879 The length of the curve is defined as the limit of the sum of the chords as the number of points of division increases without limit in such a way that at the same time each chord separately approaches zero as a limit. Since this limit will also be the measure of the length of some straight line, the finding of the length of a curve is also called "the rectification of the curve." The student has already made use of this definition for the length of a curve in his Geometry. Thus the circumference of a circle is defined as the limit of the perimeter of the inscribed (or circumscribed) regular polygon when the number of sides increases without limit. The method of the next section for finding the length of a plane curve is based on the above definition, and the student should note very carefully how it is applied. 225. Lengths of plane curves. Rectangular coordinates. We shall now proceed to express in analytical form the definition of the last section. Given the curve y =m and the points P (a, c), Q (6,d) on it ; to find the length of the arc PQ. Take any number (= n) of points on the curve between P and Q, say P\ P'\ • •, P^*\ and draw the chords PP\ P'P'\ •••, P<">e. Consider any one of these chords, P'P"^ for example, and let the coordinates of P' and P" be P' (x\ y') and P" (x' + Ax\ y' + Ay'). Then, as in § 102, p. 141, P^P" = y/(Axy + {Ay')\ or. '--Mm'^- [DiTiding inBide the radical by (Ax')* and moltiplylng oatside by Aa^.] But from the Theorem of Mean Value, (42), p. 167 (if Ay' is denoted by f{b) -f(a) and Aa;' by J - a), we get ^ =/■<->• x' <x^< x' ' 380 INTEGRAL CALCULUS Subatituting, we get P'P" = [l+/'(2-i)«]*Ax'. In the same manner we find P'P" = [!+/' (xJ'J'Ax', • • • • • The length of the inscribed broken line joining P and Q (sum of the chords) is then the sum of these expressions, and the length of the arc P^ is therefore, by definition, the limit of the sum [1 +/' (xo)']^^'' + [1 +/' {x,r]^^' + ••• + [!+/' {^JT *Az^"^ as n increases without limit Hence, if we denote the length of the arc FQ by «, we have, by § 220, p. 367, the formula for the length of the arcy >.& «=J^[l4-/'(r)^*dx, or, where -~ must be found in terms of z from the equation of the ax given curve. Sometimes it is more convenient to use y as the independent variable. To derive a formula to cover this case, we know from (88), p. 162, that rf 1 rf -^ = -— ; hence ax = —- ay, ax ax ay Substituting this value of dx in (-4), and noting that the corre- sponding y limits are c and c2, we get* the formula for the length of the arC| , "» -rKg) +']*'• dx where — in terms of y must be found from the equation of the given curve. INTEGRATION A PROCESS OF SUMMATION 381 Ex. 1. Find the length of the circle z* + y^ = i^. Solution, Differentiating, — = — . dz y Substituting in (A)^ «.M=j;'[i+?)'d. r Substituting y«=» r«-a;« from the equation of the] [circle in order to get everything in terms of x.J .*. arc BA = r \ , = r arc sin - : Hence the total length equals 2 rr. Ans. XT 2 EXAMPLES 1. Find the length of the arc of the semicubical parabola ajfl = x* from the origin to the ordinate x = 5 a. 2. Find the entire length of the hypocycloid x' + y' = a'. Ana. 335 a 27 Ana. Oo. X X 3. Rectify the catenary y = - (c«+ c~") from x = to the point (x, y). ^ a - -- Ana. -(e" — c «). 2^ 4. Find the length of one complete arch of the cycloid X = r arc vers - — V2 ry — y*. r dx y Ana. 8 r. Hint. U86(B). Here — = ^y ^2ry-y* 6. Find the length of the arc of the parabola y^ = 2jxc from the vertex to one extremity of the latus rectum. . p V2 . p , ,, . /iTv Ana. ^-r— +f log(l+v2). 6. Rectify the curve oy* = x (x - 3 a)^ from x = to x = 3 a. Ana. 2 a Vs. 7. Find the length in one quadrant of the curve (-) +(-) =1. ^"^ ^''\ ffi + ab + V Ana, a-^b 8. Find the length between x = a and x = 6 of the curve ev = Ana. log c8«-l + a-6. 9. The equations of the involute of a circle are tx = a (cos ^ + ^ sin $)y y = a (sin — cos 0), Find the length of the arc from ^ = to = ^i. Ana. \aO-^, 382 INTEGRAL CALCULUS 226. Lengths of plane curves. Polar coordinates. Formulas (A) and (B) of the last section for finding the lengths of curves whose equations are given in rectangular coordinates involved the differ- ential expressions [i.(g)-]'.»a[(|)Vx]W. In each case, if we introduce the differential of the independent variable inside the radical, they reduce to the form Let us now transform this expression into polar coordinates by means of the substitutions x = p cos 5, f/ = painO, Then dz=:z-^p sin Odd -f cos Odp^ and (iy = p cos 0d0 -f sin 0dp^ and we have [djc* + rfi/^*= [(- P sin ^d0 -f COS 0dpf 4- (p cos 0d0 + sin 0dp)^^ ^[p*d0'-\^dp^]K If the equation of the curve is p ^m. then dp==f{0)d0 = ^d0. Substituting this in the above differential expression, we get If then o and fi are the limits of the independent variable corresponding to the limits in (A) and (B), p. 308, we get the formula for the length of the arc, (A) 'a -S>^i%)i^ where p and ^ in terms of must be substituted from the equa- d0 tion of the given curve. INTEGRATION A PROCESS OF SUMMATION 883 In case it is more convenient to use p as the independent vari- able, and the equation is in the foim then d6 = ^^(fi)dp = '—dp. dp Substituting this in [p^dff^ -f rfp']* Hence, if ^j and p, are the corresponding limits of the inde- pendent variable />, we get the formula for the length of the arc, (■B) dO . •=/,''[''(g)'+>]''"- where — in terms of p must be substituted from the equation of the given curve. Ex. 1. Find the perimeter of the cardioid p = a (1 + cos 0), dp SoltUion. Here — = — a sin ^. de If we let vary from to r, the point P wiU gener- ate one half of the curve. Substituting in {A), p. 882, - = C'[ai (1 + cos ey + a2 sina e(]^d0 = a C'{2 + 2 cos tf)*(W = 2a f 'cos -de = 4a. a = 8 a. Ans, EXAMPLES 1. Find the length of the spiral of Archimedes, p = aB^ from the origin to the end oi the first revolution. . ^/r~nr^ . ^i /o . ^/ , . ^ o v Ans. ira vl + 4t2 + -log(2ir + vl + 4Ta). 2. Rectify the spiral p — tf^ from the origin to the point (p, ^). Hint. Ute(^. Aim. " Va» -f 1. a 3. Find the entire length of the curve p = o sin*- Am, 3ira 4. Find the circumference of the circle p = 2 r sin ^. Ans. 2 irr. 6. Find the length of the hyperbolic spiral pe=a from (pi, ^i) to (pj, ^2)- _ f 2 (a + Voa + pi«) k 384 INTEGRAL CALCULUS 6. Find the length of the arc of the cisBoidp = 2a tan^ sln0 from = to = An8. 2a|V6-2-V81og ^"^"^ l t V2(2 + V3)J $ WW 7. Find the length of the parabola p = a aec^- from tf = — -totf = -- 2 2 ^ Ana. 2a(8ec— + logtan— Y a Show that the entire length of the epicycloid 4(p«- a^)' = 27a*p2sin«^, a which is traced by a point on a circle of radios - rolling on a fixed circle of radius a, is 12 d. 227. Volumes of solids of revolution. Let V denote the volume of the solid generated by revolving the plane surface AMFC about the axis of X, the equation of the plane curve CFD being Let X (= 03f) take on a small increment Ax (= MN) ; then V takes on an increment A T, the volume of the disc gen- erated by the plane surface MNQP. In revolving, the two rectangles MNRP and MNQS generate cylinders having the same altitude Ax{-=^MN)y the exterior one having NQ^ and the interior one MP^ as radius of the base. The disc generated by MNPQ is evidently greater than the interior but less than the exterior cylinder. Hence irMF '£Lc<AV<TrNQ 'Ax\ or, dividing by Aa:, AF IT MP < -r- < irNQ . Ax Now let Ax approach zero as a limit ; then NQ approaches MF as a limit, and we get dV dx = vf(=7rMP); or, using differentials. dV= TT'ifdx^ INTEGRATION A PROCESS OF SUMMATION 385 which is the differential of the volume of the solid of revolution. Integrating, F= TT I f/^dx. Therefore if OA=a and ffB=zh^ the volume generated by revolv- ing ABDC about the axis of X may be calculated by means of the formula (J) where the value of y in terms of x must be substituted from the equation of the given curve. This formula is easily recalled if we consider a slice or disc of the solid between two planes perpendicular to the axis of revolu- tion as an element of the volume, and regard it as a cylinder of infinitesimal altitude dx and with a base of area iry\ hence of vol- ume iry^dx. Summing up all such slices (elements) from -4 to -B, we get the volume generated by revolving ABDC about the axis of X Similarly when F is the axis of revolution we use the formula (B) = IT j a>^dy, where the value of x in terms of y must be substituted from the equation of the given curve. Ex. 1. Find the volume generated by revolving the ellipse -^ + ^ = 1 about the axis of X. a 63 Solution. Since y* = - (a^ - a;«), and a* the required volume is twice the volume generated by OAB^ we get, substituting in (A), 2 •/© Jo a^^ 2ira6a .-. F = 3 4ira63 4 TO* To verify this result, let 6 = a. Then V = — — » the volume of a sphere, which is only a special case of the ellipsoid. When the ellipse is revolved about its major axis the solid generated is called a prolate spheroid ; when about its minor axis, an oblate spheroid. 886 INTEGRAL CALCULUS 1. Find the volume of the sphere generated by revolying the circle x^ + 1/^ = 1^ about a diameter. . ^"^y* 3 2. Find by integration the volume of the right cone generated by revolving the line joining the origin to the point (a, h) about the axis of X. yafe* 3 3. Find the volume of the cone generated by revolving the line of Ex. 2 about the axis of F. . iro«6 Ans. • 3 4. Find the volume of the paraboloid of revolution generated by revolving the arc of the parabola y^ = 4 ox between the origin and the point (zi, ^i) about its axis. Aim, 2 waz^ = -^ — ; i.e. one half of the volume of the circumscribing cylinder. 5. Show that the volumes generated by revolving y ^f about OX and OF are - and 2 w respectively. 6. Find the volume generated by revolving the arc in Ex. 4 about the axis of F. Am. — ^ = \ vxi*yi ; i.e. one fifth of the cylinderof altitude yi and radius of base Xi, 7. Find by integration the volume of the cone generated by revolving about OX that part of the line 4x — 5^ + 3 = which is intercepted between the coordi- nate axes. ^ 9 r Ans, — . 100 8. Find the volume of the spindle-shaped solid generated by revolving the hypocycloid x* + y* = a* about the axis of X, . 32 to* Ans. • 105 Scfi 9. Find the volume generated by revolving the witch y = about its asymptote OX. x" + 4 a« Ans. 4TV. 10. Find ther volume generated by revolving about OY that part of the parabola X* -f- y* = a* which is intercepted by the coordinate axis. wa^ Ans. — • 15 11. Find the volume of the torus (ring) generated by revolving the circle x2 + (y - 6)2 = a* about OX, Ans. 2 r^a^. 12. Find the volume generated by revolving one arch of the sine curve y = sinz about the axis of X, H Atis. — -• 2 13. Find the volume generated by revolving about OX the curve (x — 4 a) y* = ax (x - 3 a) between the limits x = and x = 3 a. Ans. —(15 — 16 log 2). INTEGRATION A PROCESS OF SUMMATION 387 14. Find the volume generated by revolving one arch of the cycloid « = r arc vers- — V2 ry — y* T about OX, its base. Hint. SubBtitute dx - ^ " , and limits y » 0, y » 2 r, in (A), p. 386. V2ry-y* Ana. 6tV*. X X 15. Find the volume generated by revolving the catenary y = - (e^ + e «) about the axis of X from x = Otox = &. Ana. -g-(«--e °) + -y- 16. Find the volume of the solid generated by revolving the cissoid ^ =: about its asymptote x = 2 a. 2 a « Ana, 2ir«a». 17. Given the slope of tangent to the tractrix — = ^ ; find the solid generated by revolving it about OX ^ Vo^^ya Ana, ^waK 18. Shovr that the volume of a conical cap of height a cut from the solid gen- erated by revolving the rectangular hyperbola x^ — y^ = a^ about OX equals the volume of a sphere of radius a. 19. Using the parametric equations of the hypocycloid (X = a co8« By y = a sin« ^, find the volume of the solid generated by revolving it about OX. Ana. 20. Find the volume generated by revolving one arch of the cycloid tx = a(^ — sin^), y= a(l -cos^), about its base OX. Ana. 6ir^*. Show that if the arch be revolved about OY the volume generated is 6 ir*a'. 21. Show that the volume of the egg generated by revolving the curve xV = (X - a) (X - 6) about OX is » J (« + &)log~ - 2 (6 - a) | . 22. Find the volume generated by revolving the curve x* - o%c« + aV = about OX. . 4 ira» Ana. • 15 23. Find the volume of the solid generated by revolving the curve x* 4- y' = 1 about OF. . 4ir Ana. -— • 5 388 INTEGRAL CALCULUS 228. Areas of surfaces of revolution. Let S denote the area of the shaded surface geuerated by revolving the arc CP{=«) about OX, the equation of the plane curve CPB being y=f(x). Let x(=OM) take on a small increment £^{=MN); then S takes on an increment AiS, the area of the band generated by the arc P(?(=A«). Draw the chord PQ. Let ^S' denote the area of the convex surface of the frustnim of the cone of revolution generated by the chord PC- Then A5' = ^^^ + ^^^y + ^y> -ebordPQ. rTbe Uitenl vm of the f milmm at ■ cone of reTolntion li equftl to one luin Lthe gum ot the clrc'unifereiiMa ot iti buea niulllplled bjr the lUnt height.] .(,.f). Multiplying and dividing the first member by AiS', and then dividing both members by A«, we get Now let Ax approach zero as a limit. Then At and Ay also approach zero and limit /A5\ _ dS limit /A.S'\ _ limit / chord PQ \ _ Ai^Oy^Bj rf/ Ai = 0\AsJ 'A« = 0^ A» )~ Hence from (A), d»~ or, using differentials, dS = 2 Trydt, INTEGRATION A PROCESS OF SUMMATION 889 which is the differential of the area of the surface of revolution. Integrating, ^ = 2 TT I yds ; or, since from (29), p. 143, ^* = 1 + ( "t^ ) ^^j we have 2-1 4 ^=^'/^D-(f)> Hence if 0-4 = a and OB = 6, the surface generated by revolving the arc CD about OX is given by the formula w> «=="r«'[i+(g)']''^. 6*2/ where the value of y and -y- in terms of x must be substituted dx from the equation of the given curve. This formula is easily remembered if we consider a narrow band of the surface included between two planes perpendicular to the axis of revolution as the element of area, and regard it as the convex surface of a truncated cone of revolution of infinitesimal slant height ds and with a middle section whose circumference equals 2 Try, hence of area 2 iryds. SuiAming up all such bands (elements) from A to By we get the area of the surface generated by revolving the arc CPD about OX. Similarly when OF is the axis of revolution we use the formula ,c «=2,jrv[i+(g)-]W where the value of x and -^ in terms of y must be substituted dx from the equation of the given curve. Ex. 1. Find the area of the surface of revolution generated by revolving the hypocycloid x' 4- y' = a' about the axis of X. Solution. Here ^ = - ?^ , y = (a* - x')l dx ri 190 INTEGRAL CALCULUS SubatiCating lu (B), p. S80, noting that the arc BA geoerates onlf one half of tha surface, we get = 2-j;"(a'-*')*(^)'dx _ 6Ttt' 6 12 *a* 1. find the area of the mirface of the sphere generated b; revolving the circle !C« + j/« = H about a diameter. Ana. irt*. 2. Find the area of the surface generated by revolving the parabola y* = 4 ox ■bout OX, from the origin to the point where x — Za. a ^ i 3. Find by iotegraUon the area of the surface of the cone generated by revolr- ing the line joining the origin Ui the point (a, b) about OX. ^^^ ,^ Va* + 6». 4. Find the surface generated by revolving the catenarj y = - {^ +e ') about 0¥ltomx = OUtx = a. ^ Am. 2wa'(l -«-'). 5. Find the surface of the .prolate spheroid generated by revolving the ellipse V' = (1 - e^ (a' - X*) about OX, Am. 2 t63 4 - 6. Find the surface of the torus (ring) generated by revolving the circle a!» + (V - 6)* = o> about OX. Aw. i w^ab. /Tinl. tJ>LDg (he positive value of Voi-i" gives the ontaide lorfaw, and the negatiTe vilne the lulde mrf ace. 7. Find the surface generated by revolving an arch of tlie cycloid a: = rar^versJ-V2ry-i,' „^ about 1(8 base. Ant. 3 8. Find the surface generated by the cycloid when revolved about the tangent at iU vertex. . S2 vr» A^. -^. 9. Find the surface of the oblate spheroid generated by revolving the ellipse nV + Wt* = o^* about its minor axis. , „ . 6» , 1 + e ' Ant. 2ira' + ir — log-^!^. Hinl. (^eawDtrlclt^ofelllpBe. ' '~' INTEGRATION A PROCESS OF SUMMATION 891 10. Find the surface generated when the cycloid is revolved about its axis. Ana. 8»f«(«--J). 11. Find the surface generated by revolving about OX that portion of the curve y = e* which is to the left of the axis of F. ^j^g^ ■-['v^-f lo^(l + V2)]. 12. A quadrant of a circle of radius a revolves about the tangent at one extrem ity ; prove that the area of the curved surface generated is ir (r — 2) aK 13. Find the surface generated by revolving the cardioid X = a (2 COB $ — cos 2 0)^ » = a(28in»-8in2»), ^^^ about OX. Ans. — - — . 6 14. Show that the surface generated by revolving the cycloid ix = a(^-sin^), y = a(l — cos^), about OX is . 3 16. Show that the surface generated by revolving the curve x* — ah^ + 8 oV = about OX from x = Otox = ais-^ C"(Sa^ - 2x«)dx = -ira*. 16. Show that the surface generated by revolving x* + 8 = 6 xy about OT from x = ltox = 2iBir( hlog2^j and that when revolved about OX it ^-rjr* 17. Show that the surface generated by revolving the cubical parabola y = x* about OXfromx = 0tox = lis2ir f Vl + 9x*x«dx = ^ ( VlOOO - 1). la Show that if we rotate i/« + 4x = 2 logy about OX from y = 1 to y = 2, the surface generated is — — -. CHAPTER XXXI SUCCESSIVE AND PARTIAL INTEGRATION 229. Successive Integration. Corresponding to succe89ive differ- entiation in the Differential Calculus we have the inverse process of ruceessive integration in the Integral Calculus. We shall illus- trate by means of examples the details of this process, and show how problems arise where it is necessary to apply it. Ex. 1. Given — ^ = 6x; to find y. Solution. We may write this 4-) ' = Ox, dx <i(g) = e«fa. Integrating, d^"^}^ ^^^' or, ^ = 3x* + ci. This may also be written dz' it) ^dx „ . — = 3x« + ci, dx or, d(^) = (3x2 + ci)(fx. Integrating again, -^ = 1(3 x\-^ ci) dx, or, dx "^ (-4) J? = X8 + CiX + C2. dx Again, dy = (x' + CiX + Cj) dx, and integrating, / «v X* CiX' (B) y = + 4. Cax + Ca. Arts. 4 2 The result (A) is also written in the form ■892 SUCCESSIVE AND PARTIAL INTEGRATION 393 and is called a double integral^ while (J5) is -written in the form ^ = I I I 6 xdxdxdx (or = I I I 6 xda?)^ and is called a triple integral. In general, a multiple integral requires two or more successive integrations. As before, if there are no limits assigned, as in the above example, the integral is indefinite; if there are limits assigned for each successive inte- gration, the integral is definite. Ex. 2. Find the equation of a curve for every point of which the second deriva- tive of the ordinate with respect to the abscissa equals 4. Solviion, Here — ^ = 4. Integrating as in Ex. 1, (D) y = 2 x^ + CiX 4 Cj. Ans. This is the equation of a parabola with its axis parallel to OY and extending upward. By giving the arbitrary constants of integration Cx and 0% all possible values we obtain all such parabolas. In order to determine Ci and Cs, two more conditions are necessary. Suppose we say (a) that at the point where x = 2 the slope of the tangent to the parabola is zero ; and (b) that the parabola passes through the point (2, — 1). (a) Substituting x = 2 and -^ = in (C) ox gives = 8 + Ci. Hence Ci = — 8, and (D) becomes y = 2 x^ — 8x + Ca. (b) The coordinates of (2, — 1) must satisfy this equation ; therefore — 1 = 8 — 16 + Ca, or, Ca = + 7. Therefore the equation of the particular parabola which satisfies all three condi- tions is y = 2xa-8x + 7. EXAMPLES d'l/ QX^ CiX^ 1. Given -^ = ax« ; find y. Ans. y = -— + -7- 4- CaX + Cs. dx8 60 2 2. Given — ^ = ; find y. Ana. y = — — + Cjx + Cs. dx' 2 2 dx' CiX^ 3. Given d«y = ; find y. Ana. y = logx + —- + CjX + cs- x" 2 4. Given — = sin ^ ; find p. Ana. p = cos ^ + -|- + Ca^ + Cg. (W» 2 894 INTEGRAL CALCULUS 5. Given 3^ = ^^ - p J ^^^^' ^^' * = ^5 " o ^^^^ + ^ + ^«^ + ^* 6. Given <Pp = sin^ co8^<pdiffl\ find p. Ans, p = ^lILt - _ gin + ci4> + cj. 6 3 7. Determine the equations of all curves having zero curvature. SinL ^=0, from(38),p. 161,8lno©ir=0. djfl Ana, y = cix^ cs, a doubly infinite system of straight lines. 8. The acceleration of a moving point is constant and equal to /; find the dis- tance (space) traversed. Hint. ftomKx.28.p.lM,^'-/. AnS. 8 = ^ + C^t + C 9. Show in Ex. 8 that Ci stands for the initial velocity and Ct for the initial distance. 10. Find the equation of the curve at each point of which the second derivative of the ordinate with respect to the abscissa is four times the abscissa, and which passes through the origin and the point (2, 4). Ans. 3y = 2z(2* — 1). d*y Cix' Mt^ 11. Given —^ = xcosx; findy. -4»w. y=xcosx— 4sinx + ---H hc»x + c«- (2x* 6 2 «A r,, d'y , . « , . 7cosx cos*x . CiX* . 12. Given -~ = sin«x ; findy. Ana, y = — — - + -^ +C2r +C8. (IX V «7 Jt 230. Partial Integration. Corresponding to partial differentior lion in the Differential Calculus we have the inverse process of partial integration in the Integral Calculus. . As may be inferred from the connection, partial integration means that, having given a differential expression involving two or more independent vari- ables, we integrate it, considering first a single one only as varying and all the rest constant. Then we integrate the result, considering another one as varying and the others constant, and so on. Such integrals are called double^ triple^ etc., according to the number of variables, and are called multiple integrals.* Thus the expression u=j jf{x, y)dydx indicates that we wish to find a function t^ of 2; and y such that * The integrals of the same name in the last section are siMCial cases of these, namely, when we integrate with respect to the same variable throughout. SUCCESSIVE AND PARTIAL INTEGRATION 395 In the solution of this problem the only new feature is that the constant of integration has a new form. We shall illustrate this by means of examples. Thus, suppose we wish to find u^ having given P = 2a: + y + 3. OX Integrating this with respect to x, considering y as constant, we have ^ o j. where ^ denotes the constant of integration. But since y was regarded as constant during this integration, it may happen that <f> involves y in some way ; in fact ^ will in general be a function of y. We shall then indicate this dependence of ^ on y by replacing <f> by the symbol ^(y). Hence the most general form of u is where ^(y) denotes an arbitrary function of if. As another problem let us find (A) u=ff{2^ + }/')dydx. This means that we wish to find u, having given — — ^-\-y* dxdy Integrating first with respect to y, regarding x as constant, we get . where '^(2:) is an arbitrary function of x and is to be regarded as the constant of integration. Now integrating this result with respect to a;, regarding y as constant, we have where <E> (y ) is the constant of integration, and ^(2:)= Cylr{x)dx. 396 INTEGRAL CALCULUS 231. Definite double integral. Geometric interpretation. Let /(^i y) be a continuous and single-valued function of x and y. Geometrically, {A) z =f{x, y) is the equation of a surface, as KL. Take some area S in the XY plane and construct upon iS^ as a base, the right cylinder whose elements are accordingly parallel to OZ, Let this cylinder inter- sect KL in the area ^S^, and now let us find the volume V of the solid bounded by S^ S\ and the cylindrical surface. We proceed as follows. At equal distances apart (= Aa;) in the area S draw a set of lines parallel to OF, and then a second set parallel to OX at equal distances apart (= Ay). Through these lines pass planes parallel to YOZ and XOZ respectively. Then within the areas S and S we have a network of lines, as in the figure, that in S being composed of rectangles, each of area Ax-^y. This construction divides the cylinder into a number of vertical columns, such as MNFQj whose upper and lower bases are corresponding portions of the networks in S* and S respectively. As the upper bases of these columns are curvilinear, we of course cannot calculate the volume of the columns directly. Let us replace these columns by prisms whose upper bases are found thus : each column is cut through by a plane parallel to XY passed through that vertex SUCCESSIVE AND PARTIAL INTEGRATION 897 of the upper base for which x and y have the least numerical values. Thus the column MNPQ is replaced by the right prism MNPR^ the upper base being in a plane through P parallel to the XOY plane. If the coordinates of P are (a:, y, z), then MP = z =/(a:, y), and therefore {B) Volume of MNPR =/(x, y) Ay -iix. Calculating the volume of each of the other prisms formed in the same way by replacing x and y in (-B) by corresponding values, and adding the results, we obtain a volume V* approximately equal to F; that is, ((7) r=22/(^y)^y-^^5 where the double summation sign XX ^^^i^*^ ^^* there are two variables in the quantity to be summed up. If now in the figure we increase the number of divisions of the network in S indefinitely by letting Aa: and Ay diminish indefinitely, and calculate in each case the double sum ((7), then obviously F' will approach F as a limit, and hence we have the fundamental result limit » . (D) F=Ay = oXX/(^' y)Ay.A2;. Let us see how to calculate this double limit. We commence by calculating ((7) for all the prisms of a row parallel to YOZ^ q?lj along the line DG. This will give us, approximately of course, the volume of a slice of F bounded by planes through P and Q parallel to the YOZ plane. To do this analytically, we sum up in ((7), keeping X constant (= 02>). Since in this summation Ax is also constant, we may write {C) in the form (^) F' = XA2:.X/(2r,y)Ay. Hence (2>) becomes limit . .. (F) r= A» = y Ax . y /(x, y) Ay. 398 INTEGRAL CALCULUS In (J?), the limits for the second sign of summation are the extreme values of y for the vertices of the network along the line i>&, and for the first sign of summation the extreme values of 2; in the entire network. Hence it should now be intuitianaUy evident that (F) becomes* dxi f{x,y)dy, HA ^ np for we have merely replaced the signs of summation by integral signs, and the limits by the values taken from the region S itself. We have accordingly the fundamental result, limit _^^^ r^^ r^<^ (E) F=Ay = oVV/(z,y)Ay.Aa:= j j f{x,y)dydx, the second integration sign applying to y and being performed first, X and dx being meanwhile regarded as constants. The process of evaluating (2>) might have been carried out by first adding up the columns in ((7) along a line parallel to OX, i.e. y remaining constant, afterwards summing up the resulting prisms by varying y, and finally passing to the limit as Aa: and Ay approach zero. These steps would be indicated by writing the differential expression in (F) in the form /(a:, y)dxdy and changing the limits. Summing up our line of reasoning, we may write limit . . /•«! /••*! F=Ay = oVy/(ar,y)Ay.Ax= j I f{x,y)dydx f{x, y) dxdy, where v^ and v, are in general functions of y, and u^ and u^ func- tions of a:, the second integral sign applying to the first differential and being calculated first. •A rigorooB proof of (G) is to be found In Goursat's Cour$ d*Analy»e Mathimatique, Vol. J, §123. An English translation, by Professor Uedrick, of this book is published by Giun & Company. SUCCESSIVE AND PARTIAL INTEGRATION Our result may be stated in the following fotm : The definite double integral Crf(^^y)dydx may he interpreted at that portion of the volume of a truncated right cylinder which it included between the plane XOY and the ■> z =f{x, y), the bate of the cylinder being the area bounded hy the curvet y =.Uy y=u^ x=a^, x=. u,. Similarly for the second integral. It is instructive to look upon the above process of finding the voliime of the solid as follows: Consider a column of infinitesimal base dydx and altitude z as an element of the volume. Summing up all such elements from y = DF to y = DG, x in the meanwhile being constant (say = OD), gives the volume of an indefinitely thin slice having FGJil as one face. The volume of the whole solid is then found by summing up all such slices from x = OA to a; = OB. In partial integration involving two variables the order of inte- gration denotes that the limits on the inside integral sign correspond to the variable whose differential is written inside, the differentials of the variables and their corresponding limits on the integral signs being written in the reverse order. Ex. 1. Find the volae of the definite double iut«gnl J^-^ "''"{x + v)dvdx. Solution. J C it+y)dydx 400 INTEGRAL CALCULUS Interpreting thin result geometHcally it means that we have found the Tolume of the solid of cylindrical shape ataudlng on OAB aa base and bounded at the top by the aurfacs (plane) z — z -k-y. The attention of the student is now particularly called to the maimer in which the limits do bound the base OAB, which coirespondB to the area S in the figure p. 306. Our solid here stands on a baae in the XY plane bounded by V = (line OB) i r I from y limits : V = va" - 3fl (quadrant of circle AB) ) » = (line OA) 1 x = a (line BE) \ 232. Value of a definite double integral over a region S'. In the last section we represented the definite double integral as a volume. This does not necessarily mean that every definite double integral is a volume, for the physical interpretation of the result depends on the nature of the quantities represented by x^ y, z. Thus, if Xy y, z are simply considered as the coordinates of a point in space, and nottting more, then the result is indeed a volume. In order to give the definite double integral in question an interpretation not necessarily involving the geometrical concept of volume, we observe at once that the variable z does not occur explicitly in the integral, and therefore we may confine ourselves to the XY plane. In fact, let us consider simply a region S in the XY plane, and a given function /(r, y). Then, drawing a network aa before, calculate the value of for each point of the network, and sum up, finding in this way and finally pass to the limit as Ax and A^ approach zero. This operation we call integrating the function f{x, y) over the region S, and it is denoted by the symbol }jfi?^y)dydx. S >ounded by the cu: x=aj, ■y = u^, y. If S is bounded by the curves x = a^, x= a,, y = it,, y = u^, then SUCCESSIVE AND PARTIAL INTEGRATION 401 We may state our result as follows : • To integrate a given function f{x^ y) over a given region S in the XOY plane means to calcvlate the value of limit ^ ^ Ax = yy.f(x, y) AyAx, Ay = ^ ^ as explained ahove^ and the result is equal to the definite double integral r r /(^ y) dydoi, or f C^f{x, y) dxdy, the limits being chosen so that the entire region S is covered. This process is indicated briefly by fjfi^^ y) ^y^^' In the sections which follow we shall show by numerous exam- ples how the area of the region itself may be calculated in this way, and also the moment of inertia of the region. Before attempting to apply partial integration to practical prob- lems it is best that the student should acquire by practice some ' facility in evaluating definite multiple integrals. r (a-y)xadydx = -— . Jr»8* /•" /•s^r 2/*l" r^a^ 7a26* I xdydx = — — Solution. I I xdydx = ( \xy\ dz =j(;"2xv;^3^<fa=[-|(a.-«.y];=?«' In partial integration involving three variables the order of integration is denoted in the same way as for two variables ; that is, the order of the limits on the integral signs, reading from the inside to the left, is the same as the order of the corresponding variables whose differentials are read from the inside to the right. 402 INTEGRAL CALCULUS •i /.I z*] Ex. a. Verify f f f xt/'dzdydx = ^. Solution, c f c "^^'^^yot" = c r r X *!'**']<'!"'* ~ jc f r*""*! '**<** 85 2 EXAMPLES Verify the following. 2 5. r f r^ain ^d^r = — ^ — (cos/S — cos a). Jo 3 ph ftVSt . 12. JT J Vrt-«ad«a = 6&». Ja Jv ^ ' 30 233. Plane area as a definite double integral. Rectangular coordi- nates. As a simple application of the theorem of the last section (p. 401), we shall now determine the area of the region S itself in the XO Y plane by double integration.* * Some of the ezamplefl that will be given in this and the following articles may be solved by means of a single integration by methods already explained (§§^1« ^^). The only reason in such cases for using successive integration is to familiarize the student vith a new method for solution which is sometimes the only one possible. SUCCESSIVE AND PARTIAL INTEGRATION 403 As before, draw lines parallel to OX and F at distances Lx and Ay respectively. Now take element of area = area of rectangle FQz=z£ki/- Ax^ the coordinates of P being (a;, y). Denoting by A the entire area of region Sj we have limit .^. (A) ^ = Ax = oVVAy.Aa:. We calculate this by the theorem on p. 401, setting /(ar, y) = l, and get OB /%C£ I dydx, OA %/ r.n where CD and CE are in general functions of x^ and OA and OB are constants giving the extreme values of x^ all four of these quantities being determined from the equations of the curve or curves which bound the region S. It is instructive to inteipret this double integral geometrically by referring to our figure. When we integrate first with respect to y, keeping x (=0C) constant, we are summing up all the elements in a vertical strip (as DF). Then integrating the result with respect to x means that we are summing up all such vertical strips included in the region, and this obviously gives the entire area of the region S, Or, if we change the order of integration, we have (0) OL /%m I dxdy^ QIC ^na where HG and HI are in general functions of y, and OK and OL are constants giving the extreme values of y, all four of these quan- tities being determined from the equa- tions of the curve or curves which bound the region S. Geometrically this means that we now commence by summing up all the elements in a horizontal strip (as G«7), and then find the entire area by summing up all such strips within the region. • r »— 1 ^ < ^ -k^— _\ . ^l*^ z 1 _j -- o|^ Q Jt vtr b'i I f^'' ' ' .■ ,-/ ■ "k f*J H pjl m t J Vi « \ > Y K •- — —• — • ^ •>* ^ Ix^ v c y f 404 INTEGRAL CALCULUS Corresponding to the two orders of summation (integration), the following notation and figures are sometimes used : (i>) A= j j dydx ; A= i i dxdy. S S o o Referring to the result stated on page 401, we may say: The area of any region is the value of the dovhle integral of the function f{x^ y) = 1 taken over that region. Or, also, from § 231, p. 399, The area equals numerically the volume of a right cylinder of unit height erected on the base S. Ex. 1. Calculate the area of the circle x* + y* = r* by double integration. SolvtUm. Summing up first the elements in a vertical strip, we have from (B), p. 408, pOApJiR ^ = I I dydz. Job Jus From the equation of the boundary curve (circle) .X 0B = - r, OA = r. I ^ dydz • ■ r J-V^^^^ Hence A = - 2 C'^^r'^ - x»dx = irt^. Am. When the region whose area we wish to find is symmetrical with respect to one or both of the coordinate axes, it sometimes saves us labor to calculate the area of only a part at first. In the above example we may choose our limits so as to cover only one quadrant of the circle, and then multiply the result by 4. Thus, — = I I dydx = I Vr^ — ^dx = .*. A = Trr". Ans. irr T SUCCESSIVE AND PARTIAL IKTEGRATION 405 Ex. 2. Calculate that portion of the area which lies above OX bounded by the semicubical parabola ^ = x" and the straight line y = x. Solution, Summing up first the elemenU in a hori- zontal strip, we have from (C), p. 403, o Jab From the equation of the line, ^B = y, and from the equation of the curve, AC = y^, solving each one for x. To detenu ine OD^ solve the two equations simultane- ously to find the point of intersection E, This gives the point (1, 1) ; hence OD = 1. Therefore ''=rjr'"'=r<^-«*=[¥-?]: 1. Find the area of the ellipse — f- ^ = 1 by double integration. «» fta Ans, 4 11 dydx = ir od. 2. Find by double integration the area between the straight line and a parabola with its axis along OX, each of which joins the origin and the point (a, b), I "dydx = — . •/to 6 a 3. Find by double integration the area of the rectangle formed by the coordi- nate axes and two lines through (a, b) parallel to the coordinate axes. Ans. a6. 4. Find by double integration the area of the triangle formed by OJC and the lines X = a and y = -x. a , ab Ans. — . 2 6. Find by double integration the area between the two parabolas 3 ^ = 25 x and6x« = 9y. Ans. 5. 6. Required the area in the first quadrant which lies between the parabola f/^ = ax and the circle y* = 2 ax — x^. An8, IT a» 2 a2 7. Find the area outside of the parabola y^ = 4 a (a — x) and inside of the circle y« = 4 aa - x2. 406 INTEGEAL CALCULUS 8^ Solve tha problenu on page 406 by flrat aumming up all the elementB in a horizontal BUip, ftnd theo •ummlDg up all each atripe. ^n*. Ex. 3, C C° dxdy = ab. Ei. 4, f C^dxd» = ~. 234. Plane area as a definite double integral. Polar coordinates. Suppose the equations of the curve or curves which bound the region S are given in polar coordinates. Then the region may be divided into checks bounded by mdial lines drawn from the origin, each one making the angle A0 with the next one, and concentric circles drawn with centers at the origin, the difference between each radius and the next one being Ap. We shall consider these checks as the elements of area of the region S. Let us calculate the area of one, say FQ, bounded by arcs with radii p and p + Af>. From Geometry, area of sector OSQ = ^ (/> + Ap)*£i0, area of sector OPS = J f^A0. Hence area of element FQ = ^(j> + A^)' Al? — ^ p^M = pA/)A5 + ^ Ap" ■ Ad. Then, aa before, the area of the region S will be limit _,_. , ^=4j = 05;^(pApA6' + iA?Afl) =ffpdpd9 + f^=li^P-hXX^P^0, the summation extending over the entire region, or, {A) A =j'j'pdpd$. S Here, again, the summation (integration) may be effected in two ways. SUCCESSIVE AND PAETIAL INTEGHATION 407 When we integrate first with respect to &, keeping p constant, it means that we sum up all the elements (checks) in a segment of a circular ring (as ABCD), and next integrating with respect to p. that we sum up all such rii^ within the entire region. Our limits then appear as follows : <^ ^-£'XI the angles XOA and XOB being in general fimctions of />, and OE and OF constants giving the extreme values of p. Suppose we now reverse the order of integration. Integrating first with respect to /», keeping constant, means that we sum up all the elements (checks) in a wedge-shaped strip (as GKLH). Then integrating with respect to 6, we sum up all such strips within the region S. Here XmntitXOI f%Oa I pdpdd, _wi. xoj 'foo OH and OG being in general functions of B, and the angles XOJ and .^OJ being constants giving the extreme values of 6. Corresponding to the two orders of summation (integration), the following notation and figures may be conveniently employed : (i)) . A =j'j'pd0dp, A =j'j'pdpde. These are easily remembered if we think of the elements (checks) as being rectangles with dimensions pdd and dp, and hence of area pdddp. 408 INTEGRAL CALCULUS Ex. 1. Find the area of the circle p = 2r cob $ by double integration. Solution, Summing up all the elements in a sector (as 0£), the limits are and 2r cos ^; and summing up all such sectorSi the limits are and - for the semicircle OXB. Substituting in (D), A 2 Jo Jo A = irf*. AtO, 2reot9 ,ry4 pdpde = — , or, EXAMPLES 1. In the above example find the area by integrating first with respect to $, 2. Find by double integration the entire area of the cardioid p = a (1 — cos S), 3ro« Ana. • 3. Find by double integration the entire area of the lemniscate p^^a^ cos 2 $, Am. a^ 4. Find by double integration the area of that part of the parabola ^ = a sec' - intercepted between the curve and its latus rectum. /•- /•atec"- QaS ^- ^XU" ^'^^= 3 6. Find by double integration the area between the two circles ^ = a cos B^ p = b cos $fb>a'f integrating first with respect to p, 6. Solve the last problem by first integrating with respect to fi. 235. Moment of inertia. Rectangular coordinates. Consider an element of the area of region S^ as FQ, If the coordinates of P are (r, y), the distance of P from is Var^ 4- 1/\ Mul- tiplying the area of element, i.e. ArAy, by the square of the distance of P from the origin, we have (A) (i^^f/^AyAx. SUCCESSIVE AND PARTIAL INTEGRATION 409 Then the value of limit 6x = Q^^ defines the moment of inertia of the area within the region S about the point 0, when the summation is extended over the entire region. The product (A) is then an element of the moment of inertia. Denoting this moment of inertia by /, we then have, as before, (C) J= C C(x* + y^)dydx» 8 the limits of integration being determined in the same way as for finding the area (p. 403). Ex. 1. Find I over the area bounded by the lines x = a, y = 0, y = -x. Solution. These lines bound a triangle OAB. Summing up all the elements in a vertical strip (as PQ), the y-limits are zero and - x a (found from the equation of the line OB). Summing up all such strips within the region (triangle), the x-limits are zero* and a(= OA). Hence a2 b* B(aP) = ab( — I ) . Ans. \4 12/ If we suppose the triangle to be composed of horizontal strips (as RS)^ EXAMPLES 1. Find I over the rectangle bounded by the lines z = a, y = h^ and the coordinate axes. . ' ^T^ a . o\j ^ a96 + a6» 2. Find I over the right triangle formed by the coordinate axes and the line joining the pomts (a, 0), (0, 6). h(a-x) 3. Find I for the region within the circle z^ :f- y* = r*. Ans. irr* • From the reflult, p. 401, we may say that tJie mommt of inertia of the area feithin the region S is the value of the double integral of the function f (x, y)=x* + p* taken over that region. 410 INTEGRAL CALCULUS 4. Find I for the ellipse -^ + ^ =" 1- Compare reaolt nltli preceding probli Ant. '^{tfl-^-t^y 5. Find J OTer Ihe region between the Btraight line and a parabola with axis along OX, each of which joins the origin and the point (a, b). 6. Find I over the region bounded bj the paratwla y* = 4 oz, the line z + y - 6 a = 0, and OX. __ ""•■ x'i'"" " + '^"•^ + j;"x""'<"" + rt"''^ = TT' 236. Homeat of inertia. Polar coerdinates. Consider an ele- ment of the area of region S, as FQ. If the codrdinates of P are (p, 8), the distance of P from is p. The area of the element in polar coordinates was found, od p. 406, to be Multiplying this by the square of the distance of P from the otigin, ** we get {A) p'ifiApAe+i^Ap.^e). Then, in conformity with the definition of moment of inertia (=/) of the last section, we say that (B) ir=o'^^p'(pApA0 + lAp.A0) dejinet the moment of inertia about of the area within the region S. Or, passing to the limit (as under § 234, p. 406), I=jjf^d$dp, the limits of integration being determined in the same way as in finding the area (p. 407). SL'CCESSIVE AND PARTIAL INTEGRATION 4 Ex. 1. Find / over the region bounded by the circle p = 2rcoe0. SoliUioTu Summing up the elements in the triiuigalar-eliaped strip OP, i (>-limiiH are zero and 2 r cob 9 (found from the equation of the circle). Summing up all Buch Btrips, the 0-limits are and - . Hence '-fli"'"' ''*"'- amiag apt ip(u,QB), Summing up first the elementB in a ciicu- lar atrip (as QR), we have ,. fi£-"-^^ its latuB Tectum, 480* 2. Find I over the entire area of the cardioid ^ = a(l — cos e). S. ««/«"=—• 8. Find I over the area of the lemntocate p* = a> cos 2 9. , wO* An.. — . 4. Find I oTer the area bounded hj one loop of the curve p = acoB2B. 237. Geaeral method for finding the areas of sur- faces. The method given in § 228 for finding the area of a eurfnce applied only to surfaces of revolution. We shall now give a more general method. I*et (A) »=/(!, y) be the equation of the surface KL in figure, and suppose it is required to calculate the area of the re^on S' lying on the surface. 412 INTEGRAL CALCULUS Denote by S the region on the JTOF plane, which is the orthogo- nal projection of S' on that plane. Now pass planes parallel to YOZ and XOZ at common distances Ax and Ay respectively. As in § 231, these places form truncated prisms (as FB) bounded at the top by a portion (as PQ) of the given surface whose projection on .the XOY plane is a rectangle of area /ixAy (as AB)y which rectangle also forms the lower base of the prism, the coordinates of P being {qc^ y, z). Now consider the plane tangent to the surface KL at P. Evi- dently the same rectangle AB is the projection on the XOY plane of that portion of the tai^gent plane (PB) which is intercepted by the prism PB, Assuming 7 as the angle the tangent plane makes with the XOY plane, we have area AB = area PB • cos 7, rThe projection of a plane area upon a second plane is equal to the area of thel [portion projected multiplied by the cosine of the angle between the planes. J or, AyAx = area PB . cos 7. 1 But cos 7 = [-(sy-d)'] »' rCosine of angle between tuigent plane, (70) , p. 274, and XO Fl Lplane found by method given in Solid Analytic Geometry.] area PB hence Ai/Ax = [-(DHDT' or. ^.P.=[..(|)V(J;)-]W which we take as the element of area of the region /?. We then define the area of the region S' as the summation extending over the region ^, as in § 231. Denot- ing by A the area of the region S\ we have '« ^=//['+(i)*+(S)']'""-- S SUCCESSIVE AND PARTIAL INTEGRATION 418 the limits of integration depending on the prafection on the XOY plane of the region whose area we wish to calculate. Thus for (B) we chose our limits from the boundary curve or curves of the region S in the XOF plane precisely as we have been doing in the previous four sections. If it is more convenient to project the required area on the XOZ plane, use the formula "^ ^=//[>+©'+(i)']*'^ 8 where the limits are found from the boundary of the region Sy which is now the projection of the .required area on the XOZ plane. Similarly we may use ^=//['+(g)"+(S)l''-^- 8 the limits being found by projecting the required area on the YOZ plane. In some problems it is required to find the area of a portion of one surface intercepted by a second surface. In such cases the partial derivatives required for substitution in the formula should be found from the equation of the surface whose partial area is wanted. Since the limits are found by projecting the required area on one of the coordinate planes, it should be remembered that — To find the projection of the area required on the XOY plane^ eliminate z between the equations of the surfaces whose intersections form the boundary of the area. Similarly we eliminate y to find the projection on the XOZ plane^ and X to find it on the YOZ plane. This area of a surface gives a further illustration of integration of a function over a given area. Thus in (-B), p. 412, we integrate the function „ „ , over the projection on the XO Y plane of the required curvilinear surface. 414 ISTEGRAL CALCULUS Ex. 1. Find the ares iDtegiutlou. Solution. Let ABC u of the Hurface of sphere z* + v* + ** = t* by donMa the figure be one eighth of the surface o( the sphere. ■-©'-(g)'= , y»_j' + y* + i'_ -t^-V* The projection of the area required on the JOy plane iaAOB, a region bounded by z = 0, (OB) ; y = 0, (OA) ; i" + ]/> = r", (BA). Integrating first with respect to y, ne sum up all the elements along a strip (as DEFG) which is projected on the XOY plane in a strip also (as MNFG); that is, our i/-limits are zero and AfP'(=Vr> -- 1"). Then integrat- ing with respect to x sums up all such strips composing the surface ABC; that is, our z-llmits at« zero and OA (= r). Snbeiltiitlng In (-*). J _ /■■■ n^r*-!* rdydx _ Tf* 8 "J, Jo = 4«4. Am. Vj^ - js - The ct f whose b jre of radius r is on the surface of a right cylinder, Find the surface of the cylinder intercepted b; Ex.2. the radius o the sphere. Solution. Taking the origin at the center of the sphere, an element of the cylinder for the z-axis, and a diameter of a right section of the cylinder for the z-axis, the equation of the sphere is x' + v^ + z* =i i^, and of the cylinder x' + y^ = TX. ODAPB is evidently one fourth of the cylindrical surface required. Since this area projects into the semicircular arc ODA on the SOT plane, there is no region 5 from which to determine our limits in this plane ; hence we will project our area on, say, the XOZ plane. Then the region S over which we integrate is OACB, which is bounded by 1 = 0, {OA); x = 0,{OB); i' + TX = T^,(AClf); the last equation being found by eliminating y between the equations of the two surfaces. Integrating first with respect to z means that we sum up all the elements in a verti cal strip (as PD), the i-limits being zero and Vr' — rx. Then on integrating witii respect to z we sum up ail such stripe, the x-limits being Since the required surface lies on the cylinder, the partial derivatives required for formula (C), p. 413, must be found from the equation of the cylinder. SUCCESSIVE AND PARTIAL INTEGRATION 415 Hence — = , -i^ = 0. dz 2y dz Substitating in (C), p. 413, f=rj[~^['-('-iF^)T"- Substituting the value of y in terms of x from the equation of the cylinder, Jo Jo voic-a^ Jo Vrx-x^ Jo ^* EXAMPLES 1. In the preceding example find the surface of the sphere intercepted by the cylinder. ^r ^Vrx-x^ dydx 2. The axes of two equal right circular cylinders, r being the radius of their bases, intersect at right angles ; find the surface of one intercepted by the other. Hint. Take jfl + z*'=r* and a:* + y«= r« as equations of oylinders. Jo «w i/-.a _ /1.2 Vra-x2 3. Find the area of the portion of the surface of the sphere x^ -hy^ •{■ z^ = 2ry lying within the paraboloid y = ax^ -{- hz\ , 2 ttt Ans, —z^- 4. Find the surface of the cylinder x^ + y* = f* included between the plane 2 = mx and the XOY plane. Ans. 4 r^. 6. Find the surface of the cylinder z^ + (x cos a + y sin o)» = r^ which is situ- ated in the positive compartment of co5rdinates. IRift/. The axis of this cylinder Is the line 2» 0, xoofla+y8ina*0; and the radius of base is r. Ans, sm a cos a 6. The diameter of a sphere whose radius is r is the axis of a right prism with square base of side 2 a. Find the surface of the sphere intercepted by the prism. Ans. 8 r ( 2 a arc sin — — r arc sin — - — ) 7. Find the surface of the sphere x^ + y* + a" = a^ in the first octant inter- cepted between the planes x = 0, y = 0, x = 6, y = &. Atvs. a ( 2 6 arc sin — — ^^=; — a arc sin |. 238. Volumes found by triple integration. In many cases the volume of a solid bounded by surfaces whose equations are given may be calculated by means of three successive integrations, the 416 INTEGRAL CALCULUS process being merely an extension of the methods employed in the preceding sections of this chapter. Suppose the solid in question be diirided by planes parallel to the coordinate planes into rectang^ar parallelopipeds having the dimensions A2;, Ay, A2;. The volume of one of these parallelo- pipeds i8 Az.Ay.Ax, and we choose it as the element of volume. Now sum up all such elements within the region B bounded by the given surfaces by first summing up all the elements in a column parallel to one of the coordinate axes; then sum up all such columns in a slice parallel to one of the coordinate planes containing that axis, and finally sum up all such slices within the region in question. The volume V of the solid will then be the limit of this triple sum as A2, Ay, Ax each approaches zero as a limit. That is, limit the summations being extended over the entire region B bounded by the given surfaces. Or, what amounts to the same thing, F= j j j dzdj/doD, It the limits of integration depending on the equations of the bound- ing surfaces. Thus, by extension of the principle of § 232, p. 401, we speak of volume as the result of integrating the function f{Xy y, z) = 1 throughout a given region. More generally many problems require the integration of a variable function of 2:, y, and z throughout a given region, this being expressed by the notation ffjA^^ y, z) dzdydx, R which is of course the limit of a triple sum analogous to the double sums we have already discussed. The method of evaluating this triple integral is precisely analogous to that already explained for double integrals in § 232, p. 401. SUCCESSIVE AND PARTIAL INTEGRATION Ex. 1. Find the volume ol th&t portion ol ihe ellipsoid wbich lies in the Bret octant. Sotution. Let O- ABC he that portion of the ellipsoid nhose volume la required, the equations ot the bounding surfaces being 3!* tfl f <I) - + g + - = ,,(XBO. (2) * = 0, (OAB). W y = 0. (OACi, (4) 1 = 0, (OBC). PQ is an element, being one of the rectan- gular parallciopipede with dimensions Az, Ay, ix into which the planes parallel to the coordinate planes have divided the region. Inlegraling first with respect to j/r "« si™ "P *!' suc h elements in a colamn <aB flS), the *-limits being zero [from (2)] and TB = cyjl --,-^ [from (1) by solving for z]. " Iiitegrating next with respect to y, we sum np all such colum nH in a tiice (as DEMXGF), the tf-liinits being zero [from (3)1 and MQ = b-J\ - — [from equation of the curve AGB, namely ~x + ^ = ti ^y solving for y]. Lastly, Integrating with respect to z, we sum up all such slices within the entire region — ABC, the x-limit« being zero [from (4)] and OA = a. Hence ^~ i, f ""£ "" ^'^V^ -ilX""" "■"^ = Tr- Therefore the volume of the entire ellipsoid is Ex. 2. Find the volume of the solid contained between the paraboloid of revolution a:' + j/^ = az, the cylinder i" + v* = 2 ai, and the plane i = 0. Solution. The ^-limits are zero and NP{ = found by solving equation of parabolo id for z). " The i/-limit8 are zero and JlfJV(= Va m - *', found by solving equation of cylinder for y). Tbez-limitaare£eroandO.^(=2a). t' + y* 418 INTEGRAL CALCULUS The above limitB are for the solid ONAB, one half of the solid whose yolmne is required. Hence Therefore V= EXAHPLBS 1. Find the Yolome of the sphere z^ + ^^ + 2^ = v^ hy triple integration. Ajis, . 3 2. Find the volume of one of the wedges cut from the cylinder z^ + y* = v^ by the planes « = and z = mx, -r >*v^rt-a« ^mx 2 rhn Ana. 2 I I | dzdydz = . Jo Jo Jo 3 3. Find the volume of a right elliptic cylinder whose axis coincides with the z-azis and whose altitude = 20, the equation of the base being c^' + h^'^ = 6%>. Ans. 8J J 1* dzdydx = 2waJbc 4. Find the entire volume bounded by the surface (-J +{-) +(-) =!• . abe Ana. — . 90 6. Find the entire volume bounded by the surface x' + y' + «' = ai . . Ana, • 35 6. Find the volume cut from a sphere of radius a by a right circular cylinder with b as radius of base, and whose axis passes through the center of the sphere. 8 7. Find by triple integration the volume of the solid bounded by the planes z = a, y = 6, z = mx and the coordinate planes XOT and XOZ. Ana, \ mbaK 8. The center of a sphere of radius r is on the surface of a right circular cylin- der the radius of whose basis is - • Find the volume of the portion of the cylinder intercepted by the sphere. Ana, f (r — }) r". 9. Find the volume bounded by the hyperbolic paraboloid cz = xy, the XOr plane, and the planes x = ai, z = oj, y = 61, y = 62. ^ (os' — ai') (V — fti*) ^ 4c 10. Find the volume common to the two cylinders z' + y* = f^ and z* + 2;* = r^. Ana, —- — 8 SUCCESSIVE AND PARTIAL INTEGRATION 419 11. Find the volnme bounded by the plane jb = 0, the cylinder (X - a)2 + (y - 6)9 = r«, and the hyperbolic paraboloid xy = ex. Ans, 12. Find the volume of the solid bounded by the surfaces « = 0, x2 + y« = 4az, z^ + y^ = 2cx. Ans. c 3tc« 8a 13. Find the volume included by the surfaces y^-^- z^=z4az and x — z = a. Ana, 8 xa*. 14. Find the volume of the solid in the first octant bounded hy xy = az and *+y+'=<^ 4«*. (g-iog4)«.. 16. Find the volume included between the plane jb = 0, the cylinder y* = 2cx—x^, and this paraboloid az^ + 6^ = 2 2. 16. Find the entire volumes of the solids bounded by the following surfaces : (c) (x2 + ya + ««)« = 27 a^zyz. (d) (x* + y9 + 2* + ca - o2)« = 4c»(x» + y»). (e) (x» + y« + «»)» = cxyz. Ana, Ana. Ana, Ana. Ana. Ana. 36 Swabc _________ . 6 9as ■ • 2 2y«ca9. c« 3eo' T«a6c 4V2 239. Miscellaneous applications of the Integral Calculus. In § 227 it was shown how to calculate the volume of a solid of revolution by means of a single integration. Evidently we may consider a solid of revolution as generated by a moving circle of varying radius whose center lies on the axis of revolution and whose plane is per- pendicular to it. Thus in the figure the circle ACBDy whose plane is perpendicular to OXy may be supposed to generate the solid of revolution — EGFHy while its center moves from to JV, the radius MC (= y) varjring 420 INTEGRAL CALCULUS continuously with 0M{= x) in a manner determined by the equa- tion of the plane curve that is being revolved. We will now show how this idea may be extended to the calcu- lation of volumes that are not solids of revolution when it is pos- sible to express the area of parallel plane sections of the solid as a function of their distances from a fixed point. Suppose we choose sections of a solid perpendicular to OX and take the origin as our fixed point Assuming FDE as such a section, we have from (2>), p. 404, the area A= j I dzdy^ S X being regarded as constant and the limits of integration being extended over the area S^ {DFE), If the area of DEF is expressible as a function of its distance from the origin (= z), we then have (^) ^ ^dzdy^fix). S But from § 238, p. 416, the volume of the entire solid is V=CCCdzdydx = ff f f^^^yl^^^- E S Hence, substituting from {A)j we have {B) r=Jf(x)dx, where f(x) is the area of a section of the solid perpendicular to OX expressed in terms of its distance (= x) from the origin, the aj-liraits being chosen so as to extend over the entire region E occupied by the solid. Evidently the solid — ABC may be considered as being gener- ated by the continuously varying plane section DEF as ON (= x) varies from zero to OM, The following examples will further illustrate this principle. SUCCESSIVE AND PARTIAL INTEGRATION 421 Ex. 1. Calculate the volume of the ellipsoid Xa y2 jfl —+—+-=1 a!^ l^ c^ by means of a single integration. Solution, Consider a section of the ellipsoid perpendicular to OX, as ABCD with semiaxes 6' and c\ The equation of the ellipse UEJG in the XOT plane is Solving this for y (= h') in terms of X (= OM) gives a Similarly from the equation of the ellipse EFQI in the XOZ plane we get a Hence the area of the ellipse (section) ABCD is T6r=^(a3-x«) =/(«). Substituting in (B), p. 420, y=l!2: C (a2 - z'^)dx = ^iroftc. Ans. a^ J-a 3 We may then think of the ellipsoid as being generated by a variable ellipse ABCD moving from G to E^ its center always on OX and its plane perpendicular to OX, Ex. 2. Find the volume of a right conoid with circular base, the radius of base being r and altitude a. Solution. Placing the conoid as shown in the figure, consider a section PQR perpendicular to OX, This section is an isosceles triangle; and since RM=y/2rz-x^ (found by solving x^ -h y^=^2 rz^ the equation of the circle OR A Q, for y) and JfcfP = a, the area qt the section is aV2fx-x2=/(x). Substituting in (B), p. 420, F= a f V2rx-x2dx = Jo Tf^a Ans, This is one half the volume of the cylinder of the same base and altitude. 422 INTEGRAL CALCULUS Ex. 3. A rectangle moves from a fixed point, one side varying as the distance from this point, and the other aa the square of this distance. What is the volume generated while the rectangle moves a distance of 2 ft.? Ans, 4^ cu. ft. Ex. 4. On the double ordinates of the ellipse — h t:: = 1, isosceles triangles of vertical angle 90° are described in planes perpendicular to that of the ellipse. Find the volume of the solid generated by supposing such a variable triangle moving from one extremity to the other of the major axis of the ellipse. . 4 ab^ AfU. • 3 Ex. 6. Given a right circular cylinder of altitude a and radius of base r. Through a diameter of the upper base pass two planes which touch the lower base on opposite sides. Find the volume of the cylinder included between the two planes. Ana. (w — |) ca^. Ex. 6. Two cylinders of equal altitude a have a circle of radius r for their com- mon upper base. Their lower bases are tangent to each other. Find the volume common to the two cylinders. . 4 r^ 3 Ex. 7. Determine the amount of attraction exerted by a thin, straight, homo- geneous rod of uniform thickness, of length 2, and of mass if, upon a material point P of mass m situated at a distance of a from one end of the rod in its line of direction. — « — y^l Solution. Suppose the rod to be divided __, J into equal infinitesimal portions (elements) of length dx. —- = mass of a unit length of rod ; * hence —dx = mass of any element. Newton^s Law for measuring the attraction between any two masses Is • < i.A ^' product of masses force of attraction = — ; (distance between them)' therefore the force of attraction between the particle at P and an element of the rod is — mdx m (x 4- a)2 which is then an element of the force of attraction required. The total attraction between the particle at P and the rod being the limit of the sum of all such ele- ments between x = and x = 2, we have force of attraction = Mm /•? dx Mm = I = 4- - . Ans. I Jo (x + a)-» a{a-^l) SUCCESSIVE AND PARTIAL INTEGRATION 428 Ex. 8. A vessel in the form of a right circular cone is filled with water. If A is its height and r the radius of hase, what time will it require to empty itself through an orifice of area a at the vertex ? Solution. Neglecting all hurtful resistances, it is known that the velocity of discharge through an orifice is that acquired hy a body falling freely from a height equal to the depth of the water. If then x denote depth of water, . Denote by dQ the volume of water discharged in time dtf and by dx the corresponding fall of surface. The vol- ume of water discharged through the orifice in a unit of time is y a V 2 fifx, being measure d as a right cylinder of area of base a and altitude v (= V2gz). Therefore in time dt, (A) dQ = aV2gzdL Denoting by 8 the area of surface of v^ter when depth is x, we have, from Geometry, 8 x^ „ ,wr^^. But the volume of water discharged in time dt may also be considered as the volume of cylinder AB of area of base S and altitude dx ; hence irr^E^dx (E) dQ = 8dx = —j^. Equating (A) and (B) and solving for dt, Tt*c«dx dt = Therefore ah^V2gx 'i 2Tr«VJi ah*V2ffX baV2g Ana, CHAPTER XXXII ORDINART DIFFERENTIAL EQUATIONS* 240. Differential equations. Order and degree. A differential equation is an equation involving derivatives or differentials. Differential equations have been frequently employed in this book, the foUowmg being examples. (3) tmf^ = p. (4) g=12(2:r-l). (6) dy = -^dx. a'y (6) dp a"* sin 2 d dd. (7) i«y = (20«'-12a;)da?. ^ ^ '^di? ^ dxdy dx (10) a*?* a^a^az = (1 + 3 xyz + j^i^^u. Ex. 1, p. 155 (^), p. 98 Ex.1 Ex.2 Ex. 3 Ex.1 Ex.7 Ex. 8 Ex.7 p. 113 p. 146 p. 145 p. 146 p. 197 p. 207 p. 207 * A few types only of differential equatlonB are treated In this chapter, namely snch as the student is likely to encounter in elementary work in Mechanics and Physics. 424 ORDINARY DIFFERENTIAL EQUATIONS 426 In fact, all of Chapter XIII in the Differential Calculus and all of Chapter XXV in the Integral Calculus treats of differential equations. An ordinary differential equation involves only one independent variable. The first seven of the above examples are ordinary differential equations. K partial differential equation involves more than one independent variable, as (8), (9), (10). In this chapter we shall deal with ordinary differential equations only. The order of a differential equation is that of the highest deriva- tive (or differential) in it. Thus (3), (6), (6), (8) are of the fir^t order; (1), (4), (7) are of the second order; and (2), (10) are of the third order. The degree of a differential equation which is algebraic in the derivatives (or differentials) is the power of the highest derivative (or differential) in it when the equation is free from radicals and fractions. Thus all the above are examples of differential equa- tions of the first degree except (2), which is of the second degree, 241. Solutions of differential equations. Constants of integration. A solution or integral of a differential equation is a relation between the variables involved by which the equation is identically satis- fied. Thus {A) y=^i sin ^ is a solution of the differential equation (5) , g+y-o. For, differentiating (A)^ (C) _1 =_ c^ sin a:. Now, if we substitute (A) and (C) in (5), we get — (?j sin a; -f e?i sin a; = 0, showing that (A) satisfies (B) identically. Here c^ is an arbitrary constant. In the same manner (2>) y = ^a cos X 426 INTEGRAL CALCULUS may be shown to be a solution of (B) for any value of c^. The relation (jE) y = <?i 81^ 2: 4- Cj cos X is a still more general solution of (B). In fact, by giving particular values to c^^ and c^ it is seen that the solution (U) includes the solutions (A) and (2>). The arbitrary constants c^ and c^ appearing in these solutions are called constants of integration. A solution, such as {E)^ which contains a number of arbitrary essential constants equal to the order of the equation (in this case two) is called the general solution or the complete integral,* Solutions obtained therefrom by giving particular values to the constants are called particular solutions or particular integrals. The solution of a differential equation is considered as having been effected when it has been reduced to an expression involv- ing integrals, whether the actual integrations can be effected or not. 242. Verification of the solutions of differential equations. Before taking up the problem of solving differential equations it is best to further familiarize the student with what is meant by the solu- tion of a differential equation by verifying a number of given solutions. Ex. 1. Show that (1) y = c\X COB log X + C9X sin log x + x logz is a solution of the dififerential equation (2) a:2g_xg + 2y = xlogx. Solution. Differentiating (1), we get (3) J? = (c, - Ci) sin log X 4- (Cs + Ci) cos log x + log x + 1. ox ... (Py . .sinlogx . . .coslogx . 1 Substituting (1), (8), (4) in (2), we find that the equation is identically satisfied. * It is Bhown in works on Dlif erentlal Equations that the general soltttion has n arbitrary constants when the differential equation is of the nth order. ORDINARY DIFFERENTIAL EQUATIONS 427 Verify the following solntioDS of the corresponding differential equations. Differential equations Solutions \dx/ dx dx dz/ dx dx \dx/ dx 4. |[(^ + »)=«(* + y)- (2|f-«»-e)[log(x + y-l) + x-c] = 0. dv X '^ V 7. (z + y)« -^ = a». y — aarc tan ^ = c, 8. «— - y + z Vx« - y2 = 0. arc8in- = c-x, (2x X ^ dy 8in2x . 9. -^ + ycosx = . y = slnx — 1 + c«-"»*. dx 2 d^ dy ^ __ jp dx> dx 10. (l-X«)-^-X— -aV = 0. y = CiC«"«»*n* + Cj«-«*«««*»*. 243. Differential equations of the first order and of the first degree. Such an equation may be brought into the form Mdx -f Ndy = 0, in which M and N are functions of x and y. Differential equations coming under this head may be divided into the following t3rpe8. Type I. Variables separable. When the terms of a differential equation can be so arranged that it takes on the form (A) f(x)dx+F{y)dy^Q, where f{x) is a function of x alone and F{tf) is a function of y alone, the process is called separation of the variables^ and the solu- tion is obtained by direct integration. Thus, integrating {A)^ we get the general solution {B) ff(x) dx -^fFiy) dy = e, where c is an arbitrary constant. 428 INTEGRAL CALCULUS Equations which are not given in the simple fonn {A) may often be brought into that form by means of the following rule for sepa- rating the variables. First step. Clear of fractions^ and if the equation involves derivor tiveSy multiply through by the differential of the independent variable. Second step. Collect all the terms containing the same differential into a single term. If then the equation takes on the form XYdx -f X^ Y'dy = 0, where X, X' are functions of x alone^ and F, F' are functions of y alone^ it may be brought to the form {A) by dividing through by Ji ' Y. Third step. Integrate each part separately^ as in (B), Ex. 1. Solve the equation , ^ . « dx (1 + z^)xy Solution. First step. (1 + x«) xydy = (1 + y«) dx. Second step, (1 + y^)dx - x(l + z'^ydy = 0. To separate the variables we now divide by x(l + z^) (1 + ^)f giving dx ydy x(l4-x2) l + y* dx c v^y = 0. logx - i log(l 4- x2) - i log(l + y5) =C, log (1 + x3) (1 4- y2) = 2 log X - 2 C. This result may be written in more compact form if we replace — 2 C by log c, i.e. we simply give a new form to the arbitrary constant. Our solution then becomes log (1 4- x2) (1 + 2/3) = log x3 4- log c, log (1 4- x2) (14-1^) = log cx3, (1 4- x^) (1 4- y*) = cxa. Avs. Ex. 2. Solve the equation Solution, First step. axdy 4- 2 aydx = xydy. Second step. 2 aydx -\- x{a — y)dy = 0, To separate the variables we divide by xy^ 2 adz , (a — y) dy . h = U. X y ORDINARY DIFFERENTIAL EQUATIONS 429 Third tiep. *dx dy 2 a log X + a log y - y = C, alogx«y = C + y, logex2y = -+?^. a a By passing from logarithms to exponentials this result may be written in the form £+? x«y = c« «, £ I or, x*y = c« • c*». c Denoting the constant e^ by c, we get our solution in the form x*y = C€«. uins. Differential equations 1. ydx — xdy = 0. 2. (l + y)(ix-- (1 --x)dy = 0. 3. (1 + x) ydx + (1 - y)xdy = 0. 4. (x»-o«)(ly~yda; = 0. 5. (x2-yx«)^4-3^ + xya = 0. dx 6. u«(lc4-(©-a)(lu = 0. du_ 1-f m' do ~ 1 + ua" 9. dp + /> tan ^d^ = 0. 10. sin e cos 0d^ — cos d sin 0d0 = 0. 11. sec>^tan0d^ + 8ec20tan^ = O. 12. sec> e tan 0d0 + sec^ tan dd^ = 0. 13. xydx — (a 4- x) (6 4- y) dy = 0. 14. (1 + x'^)dy - Vl - y^dz = 0. 15. Vl _ x2dy + Vl - ya^x = o. 16. 3 c* tan ydx + (1 - e*) sec" ydy = 0. So2utum« y=^cz, (l + y)(l-x) = c. log xy + X - y = c. X — a • y«a = c X + a «+y . , y ^ + log^ = c. xy X « — a = C€". tt = « 4- c 1 — c« 2 1* — arc tan a = c p = c cos 6. cos = c cos $. tan 9 tan = c. 8in« e + sin2 = c. x-y = c-\- k)g(a + x)«y*. arc sin y — arc tan x = c. y Vl-x« 4- X Vl - y2 = c, tan y = c (1 — c*)8 480 INTEGRAL CALCULUS Type n. Homogeneous equations. The differential equation is said to be homogeneous when M and N are homogeneous func- tions of X and y of the same degree.* Such differential equations may be solved by making the substitution y=vx. This will give a differential equation in v and x in which the variables are separable, and hence we may follow the rule on p. 428. Ex. 1. Solve the equation dz dx SoliUUm. y^dx + (x^ - zj/) dy = 0. Since this is a homogeneous differential equation we transform it by means of the suUtituUon ^^^ Hence dy = «fa + x<fo, and our equation becomes vWdx + (x^ - vx^ {vdz + xdv) = 0, x^wix + x8(l - V) dv = 0. To separate the variables divide by ox*. This gives ^ X V dx . rdv + (lii^» = o, logx + logt) - «= C, 10geVX= C + t, vx = cc+*' = e<^-c», vx = cer. y But = - • Hence the solution is y = cc. An^ EXAMPLES Differenlial equatUma Solutions 1. (x + y)dx + xdy = 0. as« + 2xy = c. 2. (x + y)(ix-f (y-x)dy = 0. log (x« + y2)4 _ arc tan - = c. 3. xdy - ydx = Vxa + y^dx, . 1 + 2cy - c*c« = 0. 4. (8y + 10x)dx + (6y + 7x)dy = 0. (« + y)«(2 x + y)« = c. * A function of x and y Ib iaid to be homogeneotu In the yariables if the result of replacing x and y by Ax and Ay (A being arbitrary) reduces to the original function multiplied by some power of A. This power of A is called the degree of the original function. ORDINARY DIFFERENTIAL EQUATIONS 431 DifferejUial eqiuxtiona Sohttians » 6. (t-s)dt-^tdaz=zO. te* = c. m, y dy y lin- 7. X cos - • -^ = y cos - — «. xe * = c. y y y 8. X cos - (ycfcc + xdy) = y sin - (xdy — ydx). xy cos - = c. XX X Type m. Linear equations. A differential equation is said to be linear if the equation is of the first degree in the dependent variable (usually y) and its derivatives (or differentials). The linear differential equation of the first order is of the form (A) %^^y=^^ where P, Q are functions of x alone, or constants. To integrate (A), let (JB) y = uz, where 2; is a new variable and u is a function of 2; to be determined. Differentiating (JB), Substituting (0) and (B) in (A)y we get dz du ^ u-^^-z-zr-^- Puz — Q^ or, dx ax Now let us determine, if possible, the function u such that the term in z shall drop out. This means that the coefficient of z must vanish, that is, , dx Then — = - Pdx, u and log^w = — j Pda;, giving 432 INTEGRAL CALCULUS Equation (D) then becomes dz ax To find z from the last equation, substitute in it the value of u from {E) and integrate. This gives c^dz = Qef^^dx, (F) c^z =^Qef^^dx -f C. The solution of (A) is tlien found by substituting the values of M and z from (E) and {F) in (B). This gives The proof of the correctness of ( Gr) is immediately established by substitution in (A), In solving examples coming under this head the student is advised to find the solution by following the method illustrated above, rather than by using (6)^) as a formula. Ex. 1. Solve the equation ^ ' dx x + 1 ^ ' Solution, This is evidently in the linear form (A)^ where P = — and Q = (X + 1)*. x + 1 dy dz du Let y = uz ; then ;,^ = '^ t" + ^ ;p • Substituting in the given equation (1), __^ ^^- CwJ uX uX we get dz , du 2uz , ,.« u— + 2— - :; = (X + 1)«, or, dz dx l + x ^ /ox ^^ . /^W 2U \ , . ,vi dz . /du 2u V Now to determine u we place the coefBcient of z equal to zero. This gives du 2u dx 1 H- X du 2dx u 1 4- X logeU = 21og(I + x), (8) u = ci^Kd + '>" = (1 + x)8.» • Since log, II = log, c»^ Ci + «)* = log (1 + -f)» • log, e = log (1 + a:)*, i t follows that u =» (1 + x^. ORDINARY DIFFERENTIAL EQUATIONS 433 Equation (2) now becomes, since the term in z drops out, u^ = (x + l)l. ax Replacing u by its value from (3), cte = (x + l)*dx, (4) z = a^+C. Substituting (4) and (3) in ^ = uz, we get the solution y = ^ii±lL + C(x+l)2. Ann. EXAMPLES DifferenUaX eqiuUiona Solutions 1. 2-^ = (a5 + l)». 2y = (x + l)* + c(x+l)a. 2, -^ — ^ = y = cx^-{- dx z X I — a a 3. x(l - z^dy + (2x8 - l)y(ix = caMx, y = ax -\- ex Vi - x*.* . - xydx odx „ ^^i ^""T + ^'^T^' y = ax + c(l +««)*. 5. — cos £ + <sin£=l. s = sintH-c cos t. dt ds 6. — + a cost = 1 sin 2t. ' a = sin < — 1 + c€-«*n'. dt Type IV. Equations reducible to the linear form. Some equations that are not linear can be reduced to the linear form by means of a suitable transformation. One type of such equations is (A) %^^y= ^'^ where P, Q are functions of x alone, or constants. Equation {A) may be reduced to the linear form in y and z by means of* the substitution 2 = y""^^ Such a reduction, however, is not neces- sary if we employ the same method for finding the solution as that given under Type III, p. 431. Let us illustrate this by means of an example. 434 INTEGRAL CALCULUS Ex. 1. Solve the equation (1) g + ^ = alogx.y2 ax z Solution. This is evidently in the form (^1), where P = -, Q = alog», n = 2. z Let y = U2; then -r^=u— +« — • dx dx dx Substituting in (1), we get dz ^ du uz , ^, tc-- + 2-r -i = alogX' uH*. dz dx z Now to determine u we place the coefficient of z equal to zero. This gives du u ^ du ^ dz u x logu = -logx = log-, z (3^ u = l. X Since the term in z drops out, equation (2) now becomes u — = alogx-u**', dx dz — = a logx•^(^^ I dx Replacing u by its value from (3), dz , z^ — = alogx-— , dx X dz . dx -=alogx.-, l^ a(logx)a ^ ^^ z 2 (4) z=- a(logx)a + 2(7 Substituting (4) and (3) in y = U2, we get the solution 1 2 X a(logx)2 + 2(7 or, xy [a (log x)a + 2 C] + 2 = 0. -4n«. ORDINARY DIFFERENTIAL EQUATIONS 435 EXAMPLES Differential equations 1. -T^ + xy = x'y*. dx 2. ll-z^^-xy=zaxj/*. dx 3. 8ya^~ay» = x + l. dx 4. x^(xV + xy)=l. dx 6. (y logx — l)ydx = xdy. 6. y — coax -^ = y" C08X (1 — Bin x). dx 6oiufion« ir' = x2 4- 1 + ce^. y = {cVl ~x2-a)-i. yt = a 1 w' If* x[(2 l-y2)ea^c] = e». V = (ex + logx + 1) -1 • y = tan X + sec X sin 2 + c 241 Differential equations of the nth order and of the first degree. Under this head we will consider four types which are of impor- tance in elementary work. They are special cases of linear differ- ential equations^ which we defined on p. 431. Type I. The linear differential equation U) — ^ 4- P\ — 4- P'k — 4- • • • 4- l>«f/ = O. in which the coefficients p^^ p^^ "iPn ^^^ constants. The substitution of e"" for y in the first member gives This expression vanishes for all values of r which satisfy the equation (B) r" +;>ir- » 4-;>»^""' + •••+;>. = ; and therefore for each of these values of r, e"" is a solution of (A). Equation (B) is called the auxiliary/ equation of (A). We observe that the coefficients are the same in both, the exponents in (jB) corresponding to the ordMpjj^^e derivatives in (A), and y in {A) being replaced by 1. '^ ^^rts of the auxiliary equation (B) 436 INTEGRAL CALCULUS are solutions of (A). Moreover, if each one of the solutions (C) be multiplied by an arbitrary constant, the products are also found to be solutions.* And the sum of the solutions (D), namely J^ay, by substitution, be shown to be a solution of (A), Solution (H) contains n arbitrary constants and is the general solution (if the roots are all different), while (0) eive particular solutions. Case I. When the auodliary eqvMion has imaginary roots. Since imaginary roots occur in pairs, let one pair of such roots be r^ = a-{- bij r^ = a — hi, % = V— 1 The corresponding solution is y = (?/«-^*'''>* + (?je<"-*'^* = «*" 1^1(008 62? -f i sin 6a:) -f (•,(cos hx — i sin 6x) (f = e"* \ (Cy -f (?j) cos bx -f i{Cy — <jj) sin bx\^ or, y = e"*(-4 cos 6x + ^sin Jj:), where A and B are arbitrary constants. * Substitutlxig c^e''^' for y in {A), the left-hand member becomes But this vaniflhes since r^ is a root of (J9); hence Cje**!^ is a solution of (^). Similarly for the other roots. t Replacing x by ihx in Ex. 1, p. 235, gives «'"*= 1 + tox H 1 ••-.or, L2 13 Li 15 (1) • e'^'sl + + t( bx + \; 12 Li V L3 15 ;' and replacing x by - ibx gives [2 L3 Li L5 (2) e *"*=»! + \lhx- [2 Li V L3 [6 ; But, replacing x by bx in (/<), (J3), pp. 235, 236, we get (3) cos6x=l + ••. 12 Li (4) Bmbxsabx - + •••. 13 L6 Hence (1) and (2) become e**= cos bx + i sin bxj c"**** C09 bx-i Bin bx. ORDINARY DIFFERENTIAL EQUATIONS 437 Case II. When the auxiliary equation has multiple roots. Con- sider the linear differential equation of the third order where p^j p^^ p^ are constants. The corresponding auxiliary equa- tion is ((7) r* ^p^^p^r + jp, = 0. If r^ is a root of (6?), we have shown that e**** is a solution of (F), We will now show that if r^ is a double root of (0^), then ze""^* is also a solution of (F). .Replacing y in the left-hand member of (F) by a^»', we get (H) xe^^'(r,^ ^p^r^^ +p^r^ ^p,) + e'-i^(3 r^* + 2p^r, +f a)- But since r^ is a double root of (6?), and 3 r^^ -f 2p^r^ +i?a = 0. By § 82, p. 101 Hence (ff) vanishes, and ze""^^ is a solution of (F). Correspond- ing to the double root r we then have the two solutions More generally, if r^ is a multiple root of the auxiliary equation (JB), p. 435, occurring s times, then we may at once write down s distinct solutions of the differential equation {A)y p. 435, namely In case a + hi and a — hi are each multiple roots of the auxiliary equation, occurring % times, it follows that we may write down 2 % distinct solutions of the differential equation, namely c^e"^ cos hx^ c^e"" cos 6a;, c^e"^ cos Ja:, • • ., c^'^e"^ cos hx ; <j/e" sin hx^ c^xe"^ sin hxy c^^e"^ sin 6a:, • • ., c/a:*"^ e" sin hx. Our results may now be summed up in the following rule for solving differential equations of the type where p^^ jt?j, -"i p^ are constants. 438 INTEGRAL CALCULUS First step. Write daum the corresponding auxiliary equatwn Second step. Solve completely the auxiliary equation. Third step. From the roots of the auxiliary equation write dawn the corresponding particular solutions of the differential equation as follows: m Auxiliary Equation Biffbbbntial Equation (a) Each distinct reaH , ^. i i ^. \ ' y gives a particular solution e*^»*. root Tj J (6) Each distinct pair of\ , (two particular solutions imaginary roots a±bi J le'^cos bxj «*" sin bx. , ^ 4 7.. 7 . .. ( s particular solutions obtained by (c) Amultiple root occur- \ . . , . , . .. > gives < multiplying the particular solur nng s times I . , , ^ « Itions (a) or (b) by 1, x, ar, ..., af "\ Fourth step. Multiply each of the n* independent solutions by an arbitrary constant and add the results. This gives the complete solution. Ex. 1. Solve ^ - 8^ + 4y = 0. Solution. Follow above rule. First step, H-3r*H-4 = 0, auxiliary equation. Second step. Solving, the roots are — 1, 2, 2. Third step, (a) The root — 1 gives the solution c-*. (b) The double root 2 gives the two solutions e*', x^*. Fourth step. General solution is y = cic-* + cje*^ + Cjxe**. Arts. Ex.2. 8olye^-4^ + 10^-12^ + 5» = 0. OX* dx' ox* c2x Solution. Follow above rule. First step, r* - 4 r* + 10 r* - 12 r + 5 = 0, auxiliary equation. . Second step. Solving, the roots are 1, 1, 1 ± 2 1. * A check on the Accuracy of the work is found in the fact that the first three steps must giT8 n independent solutions. ORDINARY DIFFERENTIAL EQUATIONS 439 Third step, (b) The pair of imaginary roots l±2i gives the two solutions e'co8 2x, e'8in2x(a = l, 6 = 2). (c) The doable root 1 gives the two solutions e*, ztF. Fourth step. General solution is y = cie* + cisc€* + cse* cos 2* + CfC'sin 2z, or, y = (Ci + CsX + Cs cos 2 z + C4 sin 2 x) e^. Ana, EXAMPLES Differential equationa General solutions dhi 1. — ^ = 9y. y = Cie** + CjC-«*. dhi 2. -r^ + y = 0. y = ci sin jc 4- ct cos z. 3. ^ + 12y=:7^. y = cic8* + c«c** ox^ ox ^ ^_4^ + 4y = 0. y = (ci + cax) ^*. dx^ (ix 6. ^ ~ 4-?^ = 0. y = ci + Cje*' + c«e-«*. cix' ox 6, -r? + 2^-8y = 0. y = cic*^ + Cs6-*^ + C8sin2x + C4Cos2x. ox* dx* ^ d^s d^s ^da ^ ... ,, , dt^ dt* dt 8. ^ - 12^ + 27p = 0. p = cic8» + C9fir*9 ^ cje^^^ C4<r»^. JO ^ 9. 6— + 18u = 0. tt = (Cisin2i>4-CaC082t)c8«'. dv^ dv d^y d?v 10. —^ + 2n*-4 + n*y = 0. y = (ci + C2x)co8 rvL-\-(c^-\- C4X) sin nx. dx* dx^ „ d»« . . -r/ . t^ . <V3 11. — - = «. d«» a = CiC* + c s ( C3 sm h ca cos 1 . 12. ^-7— + 6fi = 0. « = Ci^« + Cae' + C8c8«. d<' dt 440 INTEGRAL CALCULUS Type n. The linear difierential equation where X is a function of x aXone^ or constant^ andp^^ J9,, •••,!?. are constants. When X=z 0, (J) reduces to (A), Type I, p. 436, The complete solution of (J) is called the complementary fumy tion of (J). Let w be the complete solution of (J"), i.e. the complementary function of (J), and v any particular solution of (J). Then Adding, we get showing that w + v is a solution* of (J). The complete solution of (I) being w -f v, we first find the com- plementary function u by placing its left-hand member equal to zero and solving the resulting equation by the rule on p. 438. To find the particular solution t; is a problem of considerable difficulty except in special cases. For the pi-oblems given in this book we may determine the particular solution v by the following method. Differentiate successively the given equation and obtain, either directly or by elimination, a new differential equation of a higher order of Type I. Solving this by the rule on p. 438, we get its complete solution containing the complementary function u already found,! and additional terms. Determining the constants of the additional terms so as to satisfy the given differential equation, we get the particular solution v. * In works on differential equations it is shown that u + r Is the complete solution, t From the method of derivation it is obvious that every solution of the original equation must also be a solution of the derived equation. ORDINARY DIFFERENTIAL EQUATIONS 441 The method will now be illustrated by means of examples. Note. The solution of the auxiliary equation of the new derived differential equation is facilitated by observing that the left-hand member of that equation is exactly divisible by the left-hand member of the auxiliary equation used in find- ing the complementary function. Ex. 1. Solve Solution. The complementary function u of (K) is the complete solution of the equation Applying the rule on p. 488, we get as the complete solution of (L) Differentiating {K) gives ^ ' dx^ dx^ dz Multiplying (K) by 2 and adding the result to {N), we get a differential equation of Type I. Solving by the rule on p. 438, we get the com- plete solution of (0) to be y = ci€f + C2«-»* + csxe-'*, or, from (M)^ y =zu + c^ier^*. We now determine cs so that c^ze-^* shall be a particular solution v of (K), Replacing y in (K) by c»xe-'*, we get C86-**(-4 + l) = ae-«*. .*. — 8 Cg = a, or, Cs = — J a. Hence a particular solution of (K) is t = — \axe-^*, and the complete solution is y =zu + v = cic* + cje-'* — J axe-**. Ex. 2. Solve (P) ^ + nay = cosax. Solution. Solving (Q) S + "'^ = "' we get the complementary function (R) tc = Ci sin TUB + Ca cos nx. 442 INTEGRAL CALCULUS DifCerentiating (P) twice, we get (5) ^+n«^ = -o»co8ax. Multiplying (P) by a* and adding the result to (8) gives The complete solution of {T) is y = Ci8innx + CscoBnx + Cssinox + Cicosox, or, y = u + Cs sin ax + C4 cos ox. Let us now determine cs and C4 so that cs sin ox + C4 cos ox shall be a solution of (P). Replacing y in (P) by Cs sin ox + C4 cos ox, we get (71*04 — a*C4) cos ax + (n*C8 — a^Cj) sin ox = cos ax. Equating the coefficients of like terms in this identity, we get n*C4 — a'c4 = 1 and rA:i — a^s = 0, or, C4 = and Cs = 0. Hence a particular solution of (P) is cos ax v = -aa and the complete solution is cos ox y = u + « = Ci sin nx + Cj coenx H EXAMPLES Differential eqiuxtions Complete solutiona I. _4-7--2. + 12y=x. y = cic8* + cac*' + -— ^. dx' dx 144 d*v dhi dh/ dy 2- 33-2-^ + 2— ^-2--^ + y = a. y = CiBinx-f C8C0SX + {c» + C4X)c*+a. ox* dx' ox^ ax d^s t 4- 1 dt" a* d^ X* 6. -4-a*y = x>. y = CiC« + c«^*» + C8sinax4-C4COBax dx* a* d^s 2 sin ^''^ 6. — i + a*a = cosax. » = ci sin ax + cj cos ax H dx^ 2a 7. ^-^-2a5^ + a«a = c«. a = (Ci + Ci<) c«' + d«a d« ' ' (a - l)a ORDINARY DIFFERENTIAL EQUATIONS .448 Differential equations Complete aoltUicna 10. ^' - 9^ + 20. = W.. . = ce*. + ce" + l^±^i±Ie... Typem. = ^^ where X is a function of x alone, or constant. To solve this type of differential equations we have the following rule from Chapter XXXI, p. 392 : Integrate n times iuccessivelt/. Each integration will introdv^e one arbitrary constant Ex. 1. Solve -4 = xe«. Solution, Integrating the first time, ^=Cze'dx, or, T^ = xe' - e* + Ci. By (A), p, 841 Integrating the second time, -^ = r«€* • dx - fc^da: +fCidz, -^ = zc« - 2 c* + Cix + C2. dx Integrating the third time, y zzCxe^dx -('2 e^dx + fCixdx +Cc^ = x«F - 3c« + — ^- + C^ + Cs, or, y = xe' — 8 e» + Cix* + cjx + c». -4n«. where F is a function of y alone. 444 INTEGRAL CALCULUS The rule for integrating this type is as follows : First step. Multiply the left-hand member by the factor ax and the right-hand member by the equivalent factor ^dy, and integrate. The integral of the left-hand member wiU be* m Second step. Extract the square root of both members^ separate the variables^ and integrate again.^ Ex. 1. Solve -4 + a^y = 0. d!ht Solution. Here t4 = — <'>% hence of 1^1)6 IV. First step. Multiplying the left-hand member by 2 — dx and the right-hand member by 2dy, we get Integrating, (^)'= " ^^ "^ ^*' Second sUp. -^ = VCi - a^y*, dx taking the positive sign of the radical. Separating the variables, we get —^=. = dx. VCi - aV Integrating, - arc sin --L: = x + Cj, or, arc sin —^ = ax + aC^. ay This is the same as --^ = sin (ox + aC^) = sin ox cos aCi + cos ax sin aCi, 81, p. 2 Vc^ ^ . . VcJ . ^ or, y = cos aCt • sm ox H sm aC^ • cob ax a a = Ci sin ox + cs cos ax. Ans. • Since d(^V=2^^rfx. <dx/ dx dx* t Each integration introduces an arbitrary constant. ORDINARY DIFFERENTIAL EQUATIONS 445 EXAMPLES Differential equaiuma 8olution8 1. ^ = x»-2co8X. y = ^ + 28ln« + cix« + cax + c». 4. — =/8intU. 9 := - ^^B\Ti rut -^ Cit -{■ c^. (te« lyn + n ^ = a^y. ax = log(y + Vya + ci) + c,, or "^^ dxa y = cie^ + cac-«*. 8 ^ = -1-. 3t = 2ai(«*-2ci)(«* + Ci)* + Cj|. 9. ^ = -i. (ci< + c»)« + a = ciy«. <Pz . /^r- , Vci^c^f + 1-1 . ^ 10. _ = c«. tV2n = Cilog +cj. d<» Vci V* + 1 + 1 11 ^ = _ * . Find L having given that -- = and « = a, when t = 0. ^~- '=>^!I("»^*"T-')-^''^^^^ /^ CHAPTER XXXm INTE6RAPH. TABLE OF INTEGRALS 245. Mechanical integration. We have seen that the determina- tion of the area bounded by a curve C whose equation is y =/(^) and the evaluation of the definite integral f{x) dx C are equivalent problems (§ 209, p. 357). Hitherto we have regarded the relation between the variables x and y as given by analytical formulas and have applied analytic methods in obtaining the integrals required. If, however, the relation between the variables is given, not analytically, but as frequently is the case in physical investigations, graphically, i.e. by a curve,* the analytic method is inapplicable unless the exact or approximate equation of the curve can be obtained. It is, how- ever, possible to determine the area bounded by a curve, whether we know its equation or not, by means of mechanical devices. We shall consider the construction theory and the use of one such device, namely the Integraph, invented by Abdank-Abakanowicz.f Before proceeding with the discussion of this machine it is neces- sary to take up the study of integral curves. 246. Integral curves. If F(x) and f{x) are two functions so related that (A) j.^F(x)=f(x), then the curve (B) y = nx) * For instance the record made by a registering thermometer, a Bteam-«nglne indicator, or by certain testing machines. t See Lea Integraphet ; la courbe intigralc et ses applications t by Abdank-Abakanowicx, Paris, 1889. 446 INTEGRAPH 447 is called an integral curve of the curve (C) y^f{x).* The name irdegrdl curve is due to the fact that from ((7) it is seen that the same relation between the functions may be expressed as follows : {J>) rf{x)dx = F{x). F{0) = Let us draw an original curve and a corresponding integral curve in such a way. as easily to compare their corresponding points. integral curve y = F{x) original curve To find an expression for the shaded portion ((/M^P') of the area under the original curve we substitute in (A), p. 371, giving area O'Af'P' =C^f(x)dz, But from (D) this becomes area O'M'P' =j^'y(x)dx = [F{x)]x=,i= ^(^^i) = AfP.t Theorem. For the same absciasa Xi, the number giving the length of the ordinate of the integral curve {B) ie the same as the nuinber thai gives the area between the original, curve, the axes, and the ordinate corresponding to this alt^scissa. * This curve l8 Bometimes called the origiruU curve. t When Xx- O'B't the positive area O'M'RP' \a represented by the maximum ordinate NR. To the right of R the area is below the axis of X and therefore negative ; consequently the ordinates of the integral curve, which represent the algehraic sum of the areas inclosed, will decrease in passing from R to 7*. The most general integral curve is of the form y«F(x) + C, . in which case the difference of the ordinates for x= and a;» X| gives the area under the original curve. In the integral curve drawn C= Fif^^O, i.e: the general integral curve is obtained if this integral curve be displaced the distance C parallel to O K. 4:18 INTEGRAL CALCULUS The Btadent shoald also observe that (a) For the same abscissa Xi, the number giving the slope of the integral carve is the same as the number giving the length of the corresponding ordinate of the original curve [from (C)]. Hence (C) is sometimes called the curve of slopes of (B). In the figure we see that at points O, JR, T, F, where the integral curve is parallel to OX, the corresponding points (X, R\ T, V on the original curve have zero ordi- nates, and corresponding to the point W the original curve is discontinuous. (b) Corresponding to points of inflection Q, jS, 17 on the integral curve we have maximum or minimum ordinates to the original curve. For example, since d /x'\ 7^ it follows that (IE) ia an integral curve of the parabola (F) " = ? y = x« ■ ■■■ I 8 Since from (F) I —dx = i) 3 T' and from (E) it is seen that indicates the number of linear units in the ordinate M^P^\ and also the number of units of area in the shaded area OMiPi. Also since from (E) -^ = — , or tan r = -^1 dx 3 3 iriPi = y. and from (F) Xi* it is seen that the same number -^ indicates the length of ordinate MiPi and the slope of the tangent at Pi'. ^ Evidently the origin is a point of inflection of the integral curve and a point with minimum ordinate on the original curve. S!47. The integraph. The theory of this instrument is exceedingly simple and depends on the relation between the given curve and a corresponding integral curve. The instrument is constructed as follows. A rectangular carriage C moves on rollers over the plane in a direction parallel to the axis of X of the curve y =/(«). Two sides of the carriage are parallel to the axis of X; the other two, of course, I)erpendicular to it Along one of these perpendicular sides moves a small carriage Ci bearing the tracing point T, and along the other a small carriage d bearing a frame F which can revolve about an axis perpendicular 'to the surface, and carries the sharp-edged disk D to the plane of which it is perpendicular. A stud Si is fixed in the carriage Ci so as to be at the same distance from the axis of X as ia the tracing point T, A second stud 8^ is set in a crossbar of the main carriage C INTEGBAFH 449 ao u to be on tbe axis of X, A split mler B joins thrae two Btud« and slIdM upon thetD. A croBshead H elidea upon this mler and is joined to tlie frame F b; a parallelogTam. Tbe essential part of ttie Instrnment con- sista of the sharp- edged disk D, wUdi moTes under premure over a smooth plane surface (paper). Tliit disk wiU not slide, and hence as it rolls mmit always move along a path the tan- gent to nbicb at every point is tbe trace of the plane of tbe disk. If now this disk U caused to move, it la evident from tbe figure that the construction of the machine insures that the plane of the diak D shall be parallel to the ruler R. But if a Is the distance between tba ordlnatea through tbe studs 8,, Si, and r is the angle made by B (and therefore also plane of disk) with tbe aiis of X, we have {A) tanr = !!; and if v- = F(KO ia thecnrre traced by the point of contac of the disk, (B) tanT = ^.» Comparing {A) and (B), a ' "''' (C) y- = ^ Jvdi = \jf{') ^ = H^')-^ That is (dropping the primes), the curve to an integral eune of the curve The feictoT - evidently fixes merely the scale to which the integral curve la drawn and does not affect \i6form. A pencil or pen is attached to the carriage Ci in order to draw the curv< V — F(x). Displacing the disk D before tracing the original curve is equivalent U changing tbe constant of integration. ay' ilu' dx dt • Slnie x-x' + d. where d = width of niBObLne. and tberetorc ,7^ = ^ ' ^ " ^ ■ t It !■ auamed tbmt the Initrnment !■ ao conalructrd that the alMctsMU ol an; two eolreipoad lug polnu of the two curve! differ 0DI7 b; b coDtlant ; hence .c l« a faoetlon of x'. 460 INTEGKAL CALCULUS 248. Integrals for reference. Following is a table of integrals for reference. In going over the subject of Integral Calculus for the first time, the student is advised to use this table sparingly, if at all. As soon as the derivation of these integrals is thoroughly understood, the table may be properly used for saving time and labor in the solution of practical problems. SOME BLSMSNTASY FORMS 1. f{du ±dv±dw±"') = ftiu ± Cdv ± Cdw ± 2. fado = afdv. jc-dx = -—^ + C, n ± 1. n + 1 'dx 3. Jdf(x)=ff(x)dx^f(x)-^C. 5. J— = log« + a. Forms containino Integral Powers of a + 6x •/ a + ox (a + 6x)» + i J o(n -f- 1) 8. Cf{x, a-\-hx)dx. Try one of the substitations, z = a + &e, xz = a + te. * /xdx 1 ^—^ = -[a + te - a log(a + te)] +C. /x^dx 1 -— — = -[i(a + te)» - 2a(a + 6x)+ a«log(o + 6x)] +C. a A- ox cp / dx -. _1 1 2 (a + bx) a 1 , a + bx (a + bx) dx (o+te) + C. 1.6, a 4- 6x 2. r_?5_=..-L+^iog Ja»(a+ew) ax o» 3. A-^, = i riog(a + te) + ^-1 + C. ^ (a + 6x)» 6" L o + teJ TABLE OF INTEGRALS 461 15. r_*L^=_^ liog^±^+c. . 16. r ^ =ir-_L,^_g ]+c. J(o + 6x)» ft^L a + te 2(a + te)aJ FoBMB coKTAiNiMo tt^ + x^ fl^ ~ X*, a + 6x", a + &^ J a^-{-x^ a a J 1 + x« Ja^-xa 2a *a-x J x* - a« 2a *x + a /dx 1 /6 r-;: = — ^^tan-^x-v/- + C, when a>0 and &>0. Ja2-6^* 2a6 ^a-bx^ 21. rx"»(a + 6x")Pdx x'"-»+i(o + 6x«)i» + » a(m-n + l) /• , .. v ^ 6(np + m + l) 6(np + m + l)J ^ ' 22. r«-(a + bx^)Pdx = ^■■^'(a + te')p any r ^ J np + m+1 np + m + lJ 23. r — * ^x^CaH- + np — 1) /• cte_ — l)o Jx'»-"(a + dx (a + 6x")i» 1 (m — n + np — 1) 6 /• dx 24. r ^- ^x"«(a + 6x (m — l)ax»"->(a + 6x»)i»-i (m — l)o ^ x'"-"(a + te»)i» (a + 6x")p 1 m — n-^-np — l r ^c an(p-l)x«-i(a + &x*»)p-i an(p - 1) J x«(a + &x»)p-i 25 r (q + &x»)Pdx __ _ (g-i- to')i»-<-* __ h(m - n -- np ~ 1) /" (a + 6x»)Pdx J x"» "" a(m — l)x*»-i a(m — 1) J x"«-" 26 r (q -f 6x")Pdx _ (g + &x»)p anp /" (a + 6x")>~'dx J x*" (np — m + l)x"«-i np — m4-l«^ x"* 27 r_?!!!^5__ - x»-* + * a(m ~n + l) r x^-^dx ' J (a + 6x»)p " 6(m - np + 1) (a + 6x»)p-i ~ 6{m - np + 1) J (a + 6x»)p 28 r ^"^fa? _ a?""*"^ m + n — np -f 1 / ■ x*Hix J (g + 6x")p ■" gn(p - 1) (g + 6x»)p-i ~ gn(p-l) J (g + 6x«)P-i' 29 f ^ = t r g . .2n _ 3\ r <^ 1 J (a« + x2)» 2 (n - 1) a" L(aa + x«)»-i ^ V (g* + xa)"-0 ' 452 INTEGRAL CALCULUS 30 r_^_-_i_r - + (2n-8)r ^ ^1. J (a^ te«)» " 2 (n - 1) a L(a + te«)— * ^ V (a + te«)— U , where z = «*. 31 r_^^^_=if_ J (a + 6x«)« " 26(n - 1) (a + te^)*-! 26(n - 1) J (a + te^)»-i* 33. f— ^L_ = ±iog-^+C. J * (a + 6x») on o + fee" Ja + bx^ 26 \ 6/ ^^ /• x%B _ 5 __ a / * (fa; J a + te« "" 6 "" 6 J a + to* OP* /• <fa5 1 I ** . /^ dx 38. f— ^ = -1-5 f_« 39 f ^ = ^ + -L f ^ * J (a + &x3)2 2a(a + 6x2) 2a J a + 6x2' Forms containing Va + 6x 40. rxV^Ttod» = - ^<^''-«^>J<^+^ +C. •/ 16 0* 41. r..vsTte<te=^<«^iii^^^±l|^>^5«±^+o. • •/ 1066* 42. r_^= = -?i^^i:^V^i:te+c. •^ V^Tte 362 ^^ r x^dx 2(8a2-4a6x + 862x2) / ;- ^ 43. ( "^^^ _,M^ ^ W a + 6x -f C. •^ VaT6x 15^ -. /• dx 1 , Va 4- 6x — Va ^ . ^ 44. I — ■ =--log — --^-^ --ifC, fora>0. •^ X V o + 6x V a V a + 6x 4- V a ^.f /• dx 2 ^ , /o 4- 6x . ^ - ^ 45. I — = tan-i^— I- hC, fora<a •^ x Va + 6x v-a ^ ~* TABLE OF INTEGRALS 453 AR C ^ — — Vg + to 6_ r dx 47. r^5±^ = 2vnto4-ar— ^=. •^ * -^x vo + te FOBKS CONTAININO VjC» + a* 48. n^a + a«)*dx = ?Vx» + a?» + ^log(x + Vxa + o«) + C. •/ 2 2 49. 0x2 + a2)«dx = 5(2x2 + ba^ Vx2 + a* + ?^log(x + Vx^ + a^) + C. 9/0 8 60. r(x» + a»)»(to = ^^±^' + J!^ r(«« + ««)'-' <te. • + 2 61. Cx (x* + a«)2dx = ^^' "*""!? ' + C. J n + 2 . rx»(x« + a2)*dx = f (2x2 ^. a2) Vx?» + a* - ^log(x + Vx* + a*) +C. •/ o 8 62 dx . f - = log(x + Vx2 + a2)+C. •^ (x2 + a2)* 63 •'(xa dx 64. /-^^ = _^==+C. •^ (x2 + a2)« o2 Vx» + a* 56. f ^'^ , ==Vx2-f a2 + C. •^ (x2 + o2)* 66. f ^'^ , =^Vx2 + a2---log(x + Vx2>fa2)>fC. •'(x2 + a2)* 2 2 *' 67. r_?!^ = -_J== + iog{x + ViMn5)+a. •^ (x2 + a2)* vx2 + a2 •^x2(x2 + a2)* «^ -^ /• dx Vx2 + a2 . 1 , a + Vx^ + o* . ^ 60. 1: = ~r- + — I log ^ — +C. 61. ( ^ ^^ =Va2 + x»-alog ^ . +C. •/ X X •/ x^ *^ 454 INTEGRAL CALCULUS Forms contaimino Va« — a* 63. r(x« - a«)*daj = ? Vx^ - a^ - -logCx + Vx^ - a^) +C7. 64. r(x« - a«)»dx = 5(2aJ« - ha^ Vx« - a« + — - log (x + Vx^ - o») + C. J n + 1 n + 1*^ «+« x(x2 - a«)«ix = ^= -4 — +C'. 67. rx«(xa _ aa)*dx = ^(2x^ - a«) Vx^ - a^ - — log(x + Vx« - a^) +0. •/ 8 8 68. f — ^— - = log (X + Vx* - a«) + C. •^ (xa - a*)* 69. r-^-^= ^^=+C. •^ {x2 - o«)' aa vx2 - o« 70. f— ?^?_= Vii3^+C. 71 . f— ^L??— = = Vi^3^ + - log (x + Vx« - a2) + C. •^(x^-a^)* 2 2 72 ' ^ . f-^^ = --=L= + iog(x+v^^^:^+c. •^ (x« - a«)' Vx* - a2 73. f_-^ = lsec-i?+C7; r_^= = 8ec-ix+a •^x(x2-a«)* « * '^ X Vx2 - 1 /dx Vx'* — a^ ^ x»(x2-a2)* ^^ dx Vx2 - a* . 1 ,x ^^ ^ 86C"" — (*«-o«)* 2a»x« 2o« a 76. f— i? ;= Tr."' + r^.8ec-'ig+C. J X X 77. I ^^ —^ = + Iog(x + vx« - o«) +0. •/ X* X TABLE OF INTEGRALS 466 FOBXB CONTAIHINO Va* — 7^, a* . ,x 78. f{a« - x«)*(i* = ? Va* - x« + ~ sin- 1 ? + C. 79. fiaa - a!?)«daj = ? {5o« -< 2«a) V^T^a 4. ?f*sin-i- +C. •/ 8 8 ct J^ n + 1 n+lJ^ •i-J-2 81. rz(a« ~ gg)«(fa = ~ ^^^ ^\^ -f C. J 11 + 2 82. fx^a* - «^**«; = |(2x» - a^) Va^ - x^ + ^sin-i? +0. •/ 8 8 a 83. f— ^ = Bln-i ? ; r_^ = sin-i x. •^ (a« - x«)* « -^ Vl-xa dx X (a* - x»)' «* Va2-x» 84. f— ^^ = -,,-,^=. + 0. •'(aa- 86. f ^'^ =-Vo«-x«+C^' •^ (a" - x*)* __ , x«dx X 00. (aa-x2)* 2 2 a /XnXX X . ,X ^ -^^ = 8in-i- + C. ^a« - x*)' Vaa - x« « ^^ /• x»'Hfx x*""* /-^ ^ . (m-l)a2/- x«-« . 88. I r = va^ - x2 + ^^ ^— I -(te. 89. r ^ =llog ^= + C. •^x{o2-xa)* ^ a + Va2 - x* dx Va2 - x" /ox V a* — X* ^ xa(a«-x2)* «*^ /dx Va* — X* 1 , X ^ = = — -^ — rlog == + C. x»(a2--x2)* 2o2x2 2a» a ^. Va* - x* J X X 93. '(aa _ x«)* , Va2 - x2 ^ -dx = 8in-i - _j. (7. x^ X a 456 INTEGRAL CALCULUS , Forms containino V2 ox — x^, V2ax + a^ 94. CV2az-z*dx = 5-=^ V2 ox - x^ + ^ver8-»- +C. J 2 2 a 96 /: dx = vera- a ./ ■ dx = ver8-*«+C. V2ax-x« aV V2x-x« 96. |x'»v2ax--x^dx = ^ — '--{■' -^^- I x*-iv2ax-x«dx. J m + 2 m + 2./ 97 98 / dx V2 ox - x« . m-1 + dx -l)aJa;m-iV2ax-x^ /x"'dx _ _ x**-^ V2qx — x^ (2m-l)o /■ x*-*dx V2ax-X> *» m J V2ax--x« xm V2 ox - X« (2m-l)ax"» (2m x»"dx 99. ■ / V2ax~x8 x* dx = - (2 ax ~ x*)* m-3 /' V2ax--x» (2m-3)ax'» (2m-3)aJ x"-i 00 . fx V2ax-x*dx = 3a« + ax-2x« ys ^ . «' i« 1— _ v2ax - x« + — vera-i — 6 2 a 01 /: dx X V2 ax - X> V2ax-x» ax + C. xdx 02. f ^^__ - = - V2ax~x« + OYere-^-+C. •^ V2ax-x2 ** /- x^dx X + 3 a ys ;; . 3 « , x . r* 03. I ■ = v2ax-x« + -a«vera-»- + 0. •^ V2 ox - x« 2 2 a 04. f-^^!? — ^dx = V2ax-x2 + a vera-i? +C. J X a 05 ■ / V2 ax -- x» xa 2V2ax-x« ^,^_,«.^ ox = vera * — hv7» X a 06. r^^2^^^»^^_(2ax-xV J x« 3ax» 07./ dx X — a 08 / (2ax-x2)* aaV2ax-x2 xdx X (2 ox - x2)* aV2ax-X'« + C. +c. 09 . /f(x, V2S^37»)dx =/f(z -f a, V5r:^)dz, when. , = x - a TABLE OF INTEGRALS 467 110 /———^— = log{x + a + V2ax + x2) +C. V2 ox + aj2 111. ri^{x, V2ax + x«)dx =fF(z - a, V2a_aa)d2, where « = x + a. Forms containino a-\-bx ±cz^ 112. f ^ - = -=l=tan-i-4^i=L+C, when 6a<4ac. J a + bx-hcx^ y/4^ac - 6^ vToc-^ ^^^' I . v . • g = , log , + C, when b^>i •/a + 6x + cx2 V6a~4ac 2cx + 6+ V6a-4ac ac. V^T7ac + 2cx-6 114. r /^ =— J Ing ^^^-^^ +J^ZJ^r. •/a + to-cxa V62 + 4ac V62T4ac-2cx + 6 116. ( ^ — = -^ Iog(2cx + 6 + 2 V^Va + 6x + cxa) + C. •^ Va + 6x + cx^ vc 116. fVa + &x + cx«dx __ 2 ex -1-6 Va + 6x + cx« - ^""^^ 1 og (2 ex + 6 + 2 Vc Va + 6x + cx^) + C. 4c 8c> 117. I ^ = = -7^Bin-^ : + C. •^ V a + 6x - ex* V c V 6^ + 4 ac 118. rVa + 6x-ex«dx = ^^Va-h6x~cxa + ^±igg8in-i ^^.^Zi + C. •^ 4c 8c» V6a+4ac 119. C ^jg_ =^ ^+^+^' -Aiog(2cx+6-H2V^cV a+to-Hcx> )4-a •^ Va+6x-|-cx« c 2c« -^^ /• xdx Va 4- 6x — ex* 6 . , 2cx — 6 ^ 120. I ^ — = — ^ + -- sin- > ^ + C. •^ Va + 6x - cx« c 2c« VP + 4ac Other ALGSBRiiic Forms 121. fx/^^t^dx = V(a + X) (6 + x) + (a - 6)log(Va + x -J- V6 + x) +C. J \6 + X 122. fx/^-^^B = V(a-x){6-Hx) + (a -h 6) sin-i \P±| -h C. •^ \6 + X \a + 123. J^^i|dx = - V(a + X) (6 - x) - (a -h 6) sin-i^^^ + C. 458 INTEGRAL CALCULUS 24. Cj}--±J^dx = - VT^^ + 8in-ix +a 26. f— =^_ = 28in-iJ^ + C. •^ V(x - a) {/3 - z) \^-o EXPOKBNTIAL AND TbIOOWOMSTRIC ForMS 26. ra«dx = -^ + C. 129. fain xdx = -coax +C. J log a •/ 27. rC(to = e' + C. 130. fcoaxdx = 8iux + C. 28. (ef^^- \-C, 131. rtanxcix = logaecx= -logco8X+C. 32. foot xdx = log ain x -f C 33. faecxdx = f = log(8ecx 4- tanx) = logtaxi( - + - )+C. J J coax \4 2/ J* /• (2x X •coaecxdx = I = log(coflecx — cotx) = log tan- +C. ^ am X 2 36. Jaec^ xdx = tan x + C. 138. jcoaec x cot xdx = — coaec x + C 36. rcoaec«xdx= -cotx + C. 139. r8in«xdx = - - -sln2x +C. 37. faec x tan xdx = aec x + C 140. jcoa^ xdx = - + - ain 2 x + C -- /• . J ain"-! x coax ^ _ i /• 41. I am* xdx = 1 I am*-* xdx. J n n ^ coa"-*X8inx n — 1 M^ r J coa»-*X8mx . n — i /• „ , 42. I coa» xdx = H I cos"-^ xdx. •/ n n J /• dx _ __ 1 cos X n — 2 /• dx Jsin^x n — l8in"-*x n— iJain'-^x r dx _ 1 a inx n — 2 /* dx J coa" X n — 1 coa'»- * x n — 1 J ( co8»-2x C08*"-*x8in" + *x m — \ A^ C «. • - J C08»"-»x8in"-^'x .m — 1/* _, i.j 46. I C08™ X am" xdx = H I coa*"-* x sin" xdac /. , 8in»->xco8* + !x . n — 1 /• , . , coa"« X ain" xdx = 1 | coa* x ain"-* xdx. 46 . ,y /• dx _ 1 1 ^m + n — 2/* dx J 8in'"xco8"x n — 1 ain"*- ^x coa" -'x n — 1 •/ aii 8in"»xco8"-*x TABLE OF INTEGRALS 459 48. Si dx 6m"*2C08^X 'C0S"*2dX m — 1 8in"«->a5C(M^-ix m + n - 2 r dx m — 1 •'Bin"«-'x< xcos"x C08* + ^X ^q rcos^xox _ co8"-^'x m -• n + :s rcofiFa J 8in"x (n — l)8in»-ix n — 1 J 8ill»- m -• n + 2 /•cofiFxdx «x" 50 /COS C08"*X(IX cos*-*x m — 1 rcoflF-*x(ix 8in»x 8iii»x (m — n) sin* - 1 x m 61. I 8inxco8"xdx = \-C, J n + 1 CO r • - -. 8in»+ix . ^ 52. I 8m"x cos X(2x = H C ^ n + 1 tan"xdx=: j tan^-^xdr +0. cot»xdx = Jcot»-»xdx + (7. EC r • J J 8in(m4-n)x , sin (m — n)x .^ 66. I sin Tnx sin nxdx = ^^ ^ + — ^^ ^-^— +C. J 2(m + n) 2(m-n) 66./ COS mx COS , sin (m + n) X . sin (m — n) X . ^ nxdx = — ^^ '— H ^ f- + C. 2(m + n) 2(m-n) Ew r . J cos (m + n) X cos (m — n) x . ^ 67. I sin TOX cos nxdx = ^^ — -^— -^! ^^ *— + C. ./ 2 (w + n) 2 (m - n) 68. r ^ =_^__ tan-if\P^ tan?) + C, when a>6. •/a + &co6X Va2 — \fi \ifa + 6 2/ 69 •/; dx 1 V6-atan- + V6 + a log — — -^ ^ :^:r^:: + C?, when a < 6. a + 6cosx V^r^a VftH^ tan ? - VdT^ 2 a tan - + & /dx 2 2 = tan-i — -f-fi whena>5. o + 6sinx Va^-fta Va^-ft^ atan? + 6~V6a-a« /dx 1 2 ; = log hC, whena<& a + 6sinx V^TT^ a tan ? + 6 + VP^T^ 2 62. r ^ = ltan-ir^^?^) + C. ^ a2cos2x + 6«sin«x a6 \ a / 63. re^^sinnxdx = ^<^^^°^-^^^"^>+C; /^ , , c*(8inx — cosx) . -, e^sin xdx = — ^^ ^ +C. 2 460 INTEGRAL CALCULUS ^aA T— * J e«*(nsiniwc + acosnx) ^ 164. \ ef^coBTUcdz = — ^ --hC: 166. fxc«»%te=^(a«-l)+C 166^ ra?»e«dx = 5!^-? ra?»-ie«yix. 167. fa^'^'T?^ = ^""^^ ? — ra»«'x"-icfe. •^ m log a m log a J 168. C^L^z= g* logg f q""^ •^ a?" (m — l)x»"-i m — lJa;"*-i 169. re^co6»xdx = ^^^"''^<^"^^^ + ^«'"^)^!^(!Lli)f^ •^ a« + n* a« + n* ^ 170. J'x'^cos oxdx = ?^ (ox sin ox + m cos ax) - ^^'""^^ fx^-^cos axdz. Logarithmic Forms « 171. jlogx<ix = xlogx-x+C. 172. Jj— = log{logx) + logx + i logax + . • .. 174. rx-logxdx='x» + iri5?^ 1 1+C. J Ln + 1 (n + l)sj 176. /^'logx(te = ?:!^-l J^dx. 176. f x«log"x<ix = ^^ log"x ^ fx^loff^-ixdx. 177. f?"^ ~ _ g"*-^^ m + 1 r x^dx J log«x (n - l)log«-ix n - 1 J ] log"-ix INDEX [The niunbers refer to the pages.] Abaolute conTergence, 227 ; Talue, 9. Acceleration, 106. Approximate formulas, 240. Archimedes, spiral of, 283. Areas of plane carves, polar coordinates, 376, 406 ; rectangular coordinates, 371, 402. Areas of surfaces, 888, 411. Asymptotes, 262. Auxiliary equation, 436. Binomial differentials, 384. Binomial Theorem, 1, 111. Cardioid, 282. Catenary, 281. Cauchy's ratio test, 224. Change of variables, 162. Circle of curvature, 188. Cissoid, 280. Computation by series of e, 237 ; of logarithms, 238 ; of ir, 238. Concave up, 123; down, 124. Conchoid of Nicomedes, 281. Conditional convergence, 227. Cone, 2. Conjugate points, 264. Constant, 11 ; absolute, 11 ; arbitrary, 11 ; numerical, 11 ; of integration, 289, 309. Continuity, 22 ; of functions, 22. Convergency, 219. Codrdinates of center of curvature, 184. Critical values, 120. Cubical parabola, 280. Curvature, center of, 183; circle of, 183 ; definition, 169 ; radius of, 169. Curve tracing, 266. Curves in space, 271. 461 Cusp, 263. Cycloid, 96, 281. Cylinder, 2. Decreasing function, 116. Definite integration, 866. Degree of differential equation, 426. Derivative, definition, 39. Derivative of arc, 141, 143. Differential coefficient, 39. Differential equations, 424. Differential of an area, 866. Differentials, 140. Differentiation, 37 ; of constant, 48 ; of exponentials, 60; of function of a function, 67 ; of implicit function, 84, 260 ; of inverse circular functions, 70 ; of inverse function, 68 ; of logarithm, * 68, 62 ; of power, 61 ; of product, 60 ; of quotient, 62; of sum, 49; of trig- onometrical functions, 67. Double point, 262 ; integration, 396. Envelopes, 208. Equiangular spiral, 288. Evolute of a curve, 187, 218. Expansion of functions, 23L Exponential curve, 284. Family of curves, 208. Fluxions, 37, 44. Folium of Descartes, 282. Formulas for reference, 1. Function, algebraic, 17; continuity of, 22 ; definition, 12 ; derivative of, 39 ; explicit, 16; explicit algebraic, 17; exponential, 17; graph of, 24; im- plicit, 83, 202 ; increasing, decreasing, 116 ; inverse, 16, 18 ; logarithmic, 17 ; 462 INDEX Function, — continued. many>Yalued, 14; of a function, 67 rational, 16; rational integral, 16 transcendental, 17 ; trigonometric, 17 uniform, one-valued, 14. Greek alphabet, 4. Helix, 272. Homogeneous differential equation, 430. Huygens* approximation, 242. Hyperbolic spiral, 283. Hypocycloid, 282. Increasing functions, 116. Increments,* 37. Indeterminate forms, 172. Infinitesimal, 21, 141. Infinity, 21. Inflection, 136. Integer, 7. Integral curves, 446; definite, 356; definition, 288; indefinite, 289. Integraph, 448. Integration, by partial fractions, 316; by parts, 341 ; by rationalization, 320 ; by transformation, 337; definition, 287. Interval of a variable, 11. Involute, 180. Jordan, 248. Laplace, 37. Leibnitz, 44; formula. 111. Lemniscate, 281. Length of curves, 378. Lima9on, 283. Limit, change of, 366; infinite, 360; interchange of, 368; of a variable, 17; of integration, 167 ; theory of, 10, 26. Linear differential equation, 431. Litmus, 283. Logarithmic curve, 284 ; spiral, 283. Logarithms, Briggs*, 240 ; common, 240 ; Naperian, 240 ; natural, 34. Maclaurin^s theorem and series, 234. Maxima and minima, 116, 136, 160, 246. Mean value, extended theorem of, 168; generalized theorem of, 172 ; theorem of, 167. Moment of inertia, 408. Multiple roots, 100. Newton, 37, 44. Node, 262. Normal, 00, 276; plane, 271, 277. Numbers, complex, 0; imaginary, 8; irrational, 8 ; real, 8. Numerical value, 0. Order of differential equations, 426. Ordinary point, 269, 273. Osculation, 263. Osgood, 220. Parabola, 281 ; cubic, 280 ; semicubical, 280 ; spiral, 284. Parameter, 11, 208. Parametric equations, 02. Partial derivatives, 103; integration, 304. Pierpont, 250. Points, conjugate, 264; end, 266; iso- lated, 264 ; of inflection, 136 ; salient, 266; singular, 250; turning, 118. Probability curve, 284. Quadratic equation, 1. Radian, 17. Radius of curvature, 162. Rates, 148. Rational fractions, 815. Real number, 8. Reciprocal spiral, 283. Reduction formulas, 344. Rollers Theorem, 166. Semicubical parabo^ 280. Series, alternating, 226 ; arithmetical, 1 ; convergent, 210; definition, 217; diver- gent, 220; geometrical, 1; infinite, 218 ; nonconvergent, 220 ; oscillating, 220 ; power, 228. Sine curve, 284. Singular points, 269. INDEX 463 Solation of difierential equations, 426. Sphere, 2. ^ Stirling, 235. Strophoid, 283. Subnormal, 09. Subtangent, 90. SuccesBiye differentiation, 109, 146, 206 ; integration, 892. Surface, area of, 888. Tangent curve, 284. Tangent, to plane curves, 89 ; to space curves, 271. Tangent line to surface, 271. Tangent plane, 278. Taylor*s Series, 232 ; theorem, 232. Test, comparison, 222. Total differentiation, 198. Trajectory, 98, 127. Transcendental function, 17. Trigonometric differentials, 303. Triple integration, 416. Vall^-Poussin, 31. Variable, definition, 11 ; dependent, 12; independent, 12. Velocity, 102. Volumes of solids, 384, 416. Witch of Agnesi, 280. To avoid fine, this book should be returned on or before the date last stamped below LIBRARY, 9C FEB 1 4 1939 IHOOL OF JUhsJms MARl M952 iAY C- 1959 f EDUCATION